Correct option is (D) Concave, convex Myopia means short-sightedness, in this, a person has a clear vision when looking at objects close to them, but distant objects will appear blurred, to correct...
Correct option is (A) active mass The rate at which a substance reacts depends upon its active mass as the rate of reaction is directly proportional to concentration of each reactant and product.
Correct option is (A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$ Oleum, or fuming sulfuric acid, is a solution of various compositions of sulfur trioxide in sulfuric acid, or sometimes more...
A line passes through the point $(2,1,-3)$ and is parallel to the vector $(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$. Find the equations of the line in vector and Cartesian forms....
and 
, internal resistance 
and 
, connected in parallel. The equivalent emf of the combination is- (A) 
(B) 
(C) 
(D) 
Answer: Option (B)$\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$
and 
are collinear, prove that 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
be three points such that area of a 
is 4 sq units, find the value of 
.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the area of a ABC having vertices 
and 
is 35 sq units.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: To Prove: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$ Formula Used: $\cos 2 A=2 \cos ^{2} A-1$ Proof: $\text { LHS }=\cos ^{-1}\left(2 x^{2}-1\right) \ldots(1)$ Let $x=\cos A \ldots$...
Answer: AD = b units and AE = a units. D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre. The general equation of a circle: (x - h)2 + (y - k)2 = r2...
![\left[\begin{array}{ll} 4 & 0 \\ 2 & 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-00e067d8b3d687152537a9d78c193c7f_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
![\[\mathbf{3},\text{ }\mathbf{3}/\mathbf{2},\text{ }{\scriptscriptstyle 3\!/\!{ }_4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c78628f6d2e7009789c7a91d13cf2805_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{3069}/\mathbf{512}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8203641da46fd10b53398293e5d5ff5b_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, Sum of the terms Sn = \[\mathbf{3069}/\mathbf{512}\] First term a = \[3\] Common ratio r \[=\text{ }\left( 3/2 \right)/3\] \[\begin{array}{*{35}{l}} =\text{...
![\[{{\mathbf{4}}^{\mathbf{th}}},\text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{th}}}~\mathbf{and}\text{ }\mathbf{1}{{\mathbf{6}}^{\mathbf{th}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3556743371f2db97c6bfc1b6f869dbf6_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, \[\begin{array}{*{35}{l}} {{a}_{4}}~=\text{ }x \\ {{a}_{10}}~=\text{ }y \\ {{a}_{16}}~=\text{ }z \\ \end{array}\] Now, we have to show that x, y and z are in...
![\[-3\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4adc8c45b73d97c997db1b8d065f32a8_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, The fourth term of a G.P. is the square of its second term \[=\text{ }{{a}_{4}}~=\text{}{{({{a}_{2}})}^{2}}\] The first term \[{{a}_{1}}~=\text{ }\text{ }3\] We...
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From the question it is given that, The geometric progression whose 4th term \[{{a}_{4}}~=\text{ }54\] The geometric progression whose 7th term \[{{a}_{7}}~=\text{ }1458\] We know that, an = arn – 1...
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From the question it is given that, First term a = \[10\] The sum of first 14 terms of an A.P. = \[1505\] 25th term = ? We know that, \[{{S}_{n}}~=\text{ }\left( n/2 \right)\text{ }\left[ 2a\text{...
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From the question it is given that, The third term of an A.P. a3 = \[5\] The ratio of its 6th term to the 10th term \[~{{a}_{6}}~:\text{ }{{a}_{10}}~=\text{ }7\text{ }:\text{ }13\] We know that,...
![\[-6,\text{ }\left( -11/2 \right)\text{ }\text{ }5,\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5f2732129820280d06d9f02b29e6543b_l3.png "Rendered by QuickLaTeX.com")
![\[-25\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5edc190d649b7ce17dfbaab9f491e99a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{2}\text{ }+\text{ }\mathbf{5}\text{ }+\text{ }\mathbf{8}\text{}+\text{ }\ldots \text{ }+\text{ }\mathbf{x}\text{ }=\text{ }\mathbf{155}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8151df050294729c7345b4b737f553b2_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, Terms of the A.P. is \[-6,\text{ }\left( -11/2 \right)\text{ }\text{ }5,\text{ }\ldots \] The first term a = \[-6\] Common difference \[\begin{array}{*{35}{l}}...
![\[\mathbf{18}\text{ }+\text{ }\mathbf{15}{\scriptscriptstyle 1\!/\!{ }_2}\text{ }+\text{ }\mathbf{13}\text{ }+\text{ }\ldots \text{ }+\text{ }\left( -\mathbf{49}{\scriptscriptstyle 1\!/\!{ }_2} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ea46e232be47532bc726d44a06d72cd3_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[18\] Common difference d = \[15{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\text{ }18\] \[\begin{array}{*{35}{l}} =\text{ }31/2\text{ }\text{ }18 \\...
![\[15\text{ }\text{ }4n\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1da67d8c2e23ac05e352545448a85c85_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, nth term is \[15\text{ }\text{ }4n\] So, \[{{a}_{n}}~=\text{ }15\text{ }\text{ }4n\] Now, we start giving values, \[1,\text{ }2,\text{ }3,\text{ }\ldots \] in the...
From the question it is given that, The angles of a quadrilateral are in A.P. Greatest angle is double of the smallest angle Let us assume the greatest angle of the quadrilateral is a + 3d, Then,...
From the question it is given that, The sum of three numbers in A.P. = \[-3\] The product of three numbers in A.P. = \[8\] Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d Now...
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The three digits numbers which are divisible by \[9\] are \[108,\text{ }117,\text{ }126,\text{ }\ldots ,\text{ }999\] Then, first term a = \[108\] Common difference = \[9\] Last term = \[999\] We...
![\[\mathbf{20},\text{ }\mathbf{19}{\scriptscriptstyle 1\!/\!{ }_4},\text{ }\mathbf{18}{\scriptscriptstyle 1\!/\!{ }_2},\text{ }\mathbf{17}{\scriptscriptstyle 3\!/\!{ }_4},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9edef3ce91c7c8c31cec2e1c8801b2ad_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[20\] Common difference d = \[19{\scriptscriptstyle 1\!/\!{ }_4}\text{ }\text{ }20\text{ }=\text{ }77/4\text{ }\text{ }20\text{ }=\text{ }\left(...
![\[{{24}^{th}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-711d5f2bda0f7dc8c0b258798e70cf3e_l3.png "Rendered by QuickLaTeX.com")
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![\[{{72}^{nd}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e50f8eba09a2ecf6ffb5e7cfbdf72f67_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, The \[{{24}^{th}}\] term of an A.P. is twice its \[{{10}^{th}}\] term = \[{{a}_{24}}~=\text{ }2{{a}_{10}}\] We have to show that, \[{{72}^{nd}}\] term is four...
![\[\mathbf{5},\text{ }\mathbf{2},\text{ }\text{ }\mathbf{1},\text{ }\text{ }\mathbf{4},\text{ }\ldots \text{ }\mathbf{is}\text{ }\text{ }\mathbf{55}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4f9dbbb87c630934bb84bdb506e216b4_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[5\] nth term = \[-55\] Common difference d = \[2\text{ }\text{ }5\text{ }=\text{ }\text{ }3\] We know that, an = a + (n – 1)d...
![\[{{3}^{rd}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4dfc6824fdd8b67401cee977db582ccf_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, \[\begin{array}{*{35}{l}} {{a}_{3}}~=\text{ }4 \\ {{a}_{9}}~=\text{ }\text{ }8 \\\end{array}\] We know that, \[{{a}_{3}}~=\text{ }a\text{ }+\text{ }2d\text{...
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From the question it is given that, \[{{a}_{19}}~=\text{ }{{19}^{th}}~\] term of an A.P. is equal to three times its \[{{6}^{th}}\] term = \[3{{a}_{6}}\] \[{{a}_{9}}~=\text{ }19\] As we know, an = a...
![\[{{17}^{th}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c536818249352a220a99c7eb64fe6f67_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, \[{{a}_{17}}~=\text{ }5\] more than twice its \[{{8}^{th}}\] term = \[2{{a}_{8}}~+\text{ }5\] \[\begin{array}{*{35}{l}} {{a}_{11}}~=\text{ }43 \\...
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From the question it is given that, \[{{a}_{8}}~=\text{ }31\] \[{{a}_{15}}\] = the \[{{15}^{th}}\] term is \[16\] more than its \[{{11}^{th}}\] term = \[{{a}_{11}}~+\text{ }16\] we know that, an = a...
![\[\mathbf{17},\text{ }\mathbf{14},\text{ }\mathbf{11},\text{ }\ldots ,\text{ }\text{ }\mathbf{40}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2fbc9e047f77761d29b709618e235523_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[17\] Common difference = \[14\text{ }\text{ }17\text{ }=\text{ }\text{ }3\] Last term l = \[-40\] \[\begin{array}{*{35}{l}} L\text{ }=\text{...
![\[\mathbf{9},\text{ }\mathbf{12},\text{ }\mathbf{15},\text{ }\mathbf{18},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a253007e4d395b0af8f4c37b8e492a59_l3.png "Rendered by QuickLaTeX.com")
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From the question, The first term a = \[9\] Then, difference d = \[12\text{ }\text{ }9\text{ }=\text{ }3\] \[\begin{array}{*{35}{l}} 15\text{ }\text{ }12\text{ }=\text{ }3 \\ 18\text{ }\text{...
![\[\mathbf{6n}\text{ }+\text{ }\mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7d50a882fd12558ef70b68fdf599a082_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, nth term is \[\mathbf{6n}\text{ }+\text{ }\mathbf{2}\] So, \[{{T}_{n}}~=\text{ }6n\text{ }+\text{ }2\] Now, we start giving values, \[1,\text{ }2,\text{ }3,\text{...
![\[\mathbf{0},\text{ }{\scriptscriptstyle 1\!/\!{ }_4},\text{ }{\scriptscriptstyle 1\!/\!{ }_2},\text{ }{\scriptscriptstyle 3\!/\!{ }_4},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-88aebe14205e70cb6dfde155d4f95454_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{5},\text{ }\mathbf{14}/\mathbf{3},\text{ }\mathbf{13}/\mathbf{3},\text{ }\mathbf{4},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-42c82d49ed1e5991026bc008ca491f2c_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = 0 Common difference \[=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }\text{ }0\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\] So, next three...
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From the question it is given that, First term a = \[-5\] Common difference d = \[-3\] Then the first four terms are = \[\text{ }5\text{ }+\text{ }\left( -3 \right)\text{ }=\text{ }-5\text{ }\text{...
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From the question it is given that, First term of a G.P. is a = \[5\] The sum of first three terms is S3 = \[31/5\] We know that, \[{{S}_{n}}~=\text{ }a({{r}^{n}}~\text{ }1)/\left( r\text{ }\text{...
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From the question it is given that, Common ratio r = \[3\] \[{{S}_{7}}~=\text{ }2186\] We know that, \[{{S}_{n}}~=\text{ }a({{r}^{n}}~\text{ }1)/\left( r\text{ }\text{ }1 \right)\]...
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From the question it is given that, First term a is = \[7\] Then, last term is = \[448\] Sum = \[889\] We know that, last term = \[~a{{r}^{n\text{ }\text{ }1}}\] \[\begin{array}{*{35}{l}}...
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From the question it is given that, Common ratio r = \[3\] Last term = \[486\] Sum of the terms = \[728\] We know that, \[{{S}_{n}}~=\text{ }a({{r}^{n}}~\text{ }1)/\left( r\text{ }\text{ }1...
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From the question it is given that, First term a = \[27\] \[{{8}^{th}}~term\text{ }{{a}_{8}}~=\text{ }1/81\] Then, \[\begin{array}{*{35}{l}} {{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}} \\...
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From the question it is given that, \[\begin{array}{*{35}{l}} {{a}_{2}}~=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2} \\ {{a}_{5}}~=\text{ }1/16 \\ \end{array}\] We know that, \[{{a}_{2}}~=\text{...
![\[\mathbf{2}/\mathbf{9}\text{ }\text{ }\mathbf{1}/\mathbf{3}\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }+\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b4f3ccfa9ce61d9a2864ef00c5cb66a6_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, Terms of G.P. is \[\mathbf{2}/\mathbf{9}\text{ }\text{ }\mathbf{1}/\mathbf{3}\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }+\text{ }\ldots \] Sum of...
![\[\mathbf{3},\text{ }{{\mathbf{3}}^{\mathbf{2}}},\text{ }{{\mathbf{3}}^{\mathbf{3}}},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f8fa292e8e7008f7c9011a3944709cd6_l3.png "Rendered by QuickLaTeX.com")
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![\[\mathbf{1},\text{ }\mathbf{4},\text{ }\mathbf{16},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3f928ca08cf4646f118cfdc56b9773ff_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, Terms of the G.P. \[\mathbf{3},\text{ }{{\mathbf{3}}^{\mathbf{2}}},\text{ }{{\mathbf{3}}^{\mathbf{3}}},\text{ }\ldots \] Sum of the terms = \[120\] The first term...
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From the question it is given that, The nth term of a G.P. \[{{T}_{n}}~=\text{ }128\] The sum of its n terms \[{{S}_{n}}~=\text{ }255\] Common ratio r = \[2\] We know that, \[{{T}_{n}}~=\text{...
![\[\mathbf{81}\text{ }\text{ }\mathbf{27}\text{ }+\text{ }\mathbf{9}\text{ }\ldots \text{ }\text{ }\mathbf{1}/\mathbf{27}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-706b80402784be5a80314a58c125e1e4_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[81\] \[\begin{array}{*{35}{l}} r\text{ }=\text{ }-27/81 \\ =\text{ }-1/3 \\ \end{array}\] Last term l =\[~-1/27\] \[\begin{array}{*{35}{l}} ...
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![\[\mathbf{1}\text{ }+\text{ }\mathbf{2}/\mathbf{3}\text{ }+\text{ }\mathbf{4}/\mathbf{9}\text{ }+\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aa9cec9e21eec238497ab169a201a92d_l3.png "Rendered by QuickLaTeX.com")
From the question, First term a = \[1\], Common ratio \[r\text{ }=\text{ }-2/3\text{ }\times \text{ }1=\text{ }-2/3\] Number of terms n = \[6\] So, \[\begin{array}{*{35}{l}} {{S}_{6}}~=\text{...
![\[\mathbf{2}\text{ }+\text{ }\mathbf{6}\text{ }+\text{ }\mathbf{18}\text{ }+\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f61f274fcbad3fc7fbb6a73d3b67b8ec_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{1}\text{ }+\text{ }\surd \mathbf{3}\text{ }+\text{ }\mathbf{3}\text{ }+\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7e597f3850306e4d18019d35a9a6ac46_l3.png "Rendered by QuickLaTeX.com")
From the question, First term a = \[2\], Common ratio r = \[6/2\text{ }=\text{ }3\] Number of terms n = \[20\] So, \[\begin{array}{*{35}{l}} {{S}_{20}}~=\text{ }a({{r}^{n}}~\text{ }1)/r\text{...
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From the question it is give that, The sum of first three terms of a A.P. is \[15\] Let us assume three numbers are a – d, a, a + d. The sum of three terms = \[a\text{ }\text{ }d\text{ }+\text{...
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From the question it is give that, The sum of first three terms of a G.P. is \[\mathbf{39}/\mathbf{10}\] The product of first three terms of a G.P. is \[1\] Let us assume that a be the first term...
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From the question it is given that, \[{{4}^{th}}\] term \[{{a}_{4}}~=\text{ }54\] \[{{7}^{th}}\] term \[{{a}_{7}}~=\text{ }1458\] \[\begin{array}{*{35}{l}} a{{r}^{3}}~=\text{ }54 \\...
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From the question, \[\begin{array}{*{35}{l}} {{({{a}^{2}}~+\text{ }2)}^{2}}~=\text{ }a({{a}^{3}}~+\text{ }10) \\ {{a}^{4}}~+\text{ }4\text{ }=\text{ }{{a}^{4}}~+\text{ }10a \\ 4{{a}^{2}}~\text{...
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From the question it is given that, \[\begin{array}{*{35}{l}} {{a}_{5}}~=\text{ }p \\ {{a}_{8}}~=\text{ }q \\ {{a}_{11}}~=\text{ }s \\ \end{array}\] Now we have to prove that,...
From the question it is given that, a4 = x a7 = y a10 = z Now we have to prove that, x, y, z are in G.P. Then, by the formula \[{{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\]...
![\[\mathbf{x}\text{ }+\text{ }\mathbf{3},\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{9}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-512e25d91a2b4ff63cf7cd04efa4aea7_l3.png "Rendered by QuickLaTeX.com")
From the question, \[\begin{array}{*{35}{l}} {{\left( x\text{ }+\text{ }3 \right)}^{2}}~=\text{ }x\left( x\text{ }+\text{ }9 \right) \\ {{x}^{2}}~+\text{ }6x\text{ }+\text{ }9\text{ }=\text{...
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From the question, \[\begin{array}{*{35}{l}} {{x}^{2}}~=\text{ }-2/7\text{ }\times \text{ }-7/2 \\ {{x}^{2}}~=\text{ }1 \\ x\text{ }=\text{ }\pm \text{ }1 \\ \end{array}\] Therefore, \[x\text{...
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From the question it is given that, First term of G.P. a = \[{\scriptscriptstyle 3\!/\!{ }_4}\] Common ratio (r) = \[2\] Last term = \[384\] Then, by the formula \[{{a}_{n}}~=\text{ }a{{r}^{n\text{...
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From the question it is given that, a8 = \[192\] and r = \[2\] Then, by the formula \[{{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\] \[\begin{array}{*{35}{l}} {{a}_{8}}~=\text{ }a{{r}^{8\text{...
![\[\mathbf{2},\text{ }\mathbf{2}\surd \mathbf{2},\text{ }\mathbf{4},\text{ }\ldots \text{ }\mathbf{is}\text{ }\mathbf{128}?\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-88e13e2c8fd74c9ef35255aa8f69a077_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{1},\text{ }\mathbf{1}/\mathbf{3},\text{ }\mathbf{1}/\mathbf{9},\text{ }\ldots \text{ }\mathbf{is}\text{ }\mathbf{1}/\mathbf{243}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ca71eb7958ecd0891ea6053495116b7a_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, Last term = 128 First term a = 2, Then, \[\begin{array}{*{35}{l}} r\text{ }=\text{ }\left( 2\surd 2 \right)\text{ }\div \text{ }\left( 2 \right) \\ r\text{...
![\[\mathbf{3},\text{ }\text{ }\mathbf{6},\text{ }\mathbf{12},\text{ }\text{ }\mathbf{24},\text{ }\ldots ,\text{ }\mathbf{12288}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c994267795992e20cd066d203934c28d_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, Last term = 12288 First term a = 3, Then, \[\begin{array}{*{35}{l}} r\text{ }=\text{ }\left( -6 \right)\text{ }\div \text{ }\left( 3 \right) \\ r\text{ }=\text{...
![\[\mathbf{5},\text{ }\mathbf{25},\text{ }\mathbf{125},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8bb5a888eb3c6e99ec00224a78f11c6b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{3}/\mathbf{2},\text{ }{\scriptscriptstyle 3\!/\!{ }_4},\text{ }\mathbf{3}/\mathbf{8},\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-95a55d395ebcb683e446c351acb5ff3a_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = 5, Then, \[\begin{array}{*{35}{l}} ~r\text{ }=\text{ }\left( 25 \right)\text{ }\div \text{ }\left( 5 \right) \\ r\text{ }=\text{ }\left( 25...
![\[\surd \mathbf{3}\text{ }+\text{ }\mathbf{1}/\surd \mathbf{3}\text{ }+\text{ }\mathbf{1}/\mathbf{3}\surd \mathbf{3}\text{ }+\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-555b8b9b3854b1f6853df01fecd123ac_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{1}/\surd \mathbf{2},\text{ }-\mathbf{2},\text{ }\mathbf{4}\surd \mathbf{2},\text{ }\text{ }\mathbf{16},\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9f001a1b43b6a1ec5e622677c1ce5717_l3.png "Rendered by QuickLaTeX.com")
From the question, First term \[a\text{ }=\text{ }\surd 3\] Then, \[\begin{array}{*{35}{l}} r\text{ }=\text{ }\left( 1/\surd 3 \right)\text{ }\div \text{ }\left( \surd 3 \right) \\ r\text{ }=\text{...
![\[\mathbf{1}/\mathbf{6},\text{ }\mathbf{1}/\mathbf{3},\text{ }\mathbf{2}/\mathbf{3},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a8043182419fbd9ac55f92e58a226841_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{3}/\mathbf{16},\text{ }-\mathbf{3}/\mathbf{8},\text{ }{\scriptscriptstyle 3\!/\!{ }_4},\text{ }-\mathbf{3}/\mathbf{2},\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ad62a1996dd4fd8785a74ff48eeb98fa_l3.png "Rendered by QuickLaTeX.com")
From the question, First term a = \[1/6\] Then, \[\begin{array}{*{35}{l}} ~r\text{ }=\text{ }\left( 1/3 \right)\text{ }\div \text{ }\left( 1/6 \right) \\ r\text{ }=\text{ }\left( 1/3 \right)\text{...
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(iii) The multiples of 9 lying between \[300\] and \[700\] are: \[306,\text{ }315,\text{ }324,\text{ }\ldots \ldots \] This form an A.P. where a = 306 and d = 9 The last term in this series is...
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(i) The two-digit natural numbers which are divisible by \[4\] are: \[4,\text{ }8,\text{ }12,\text{ }16,\text{ }\ldots ..\] This form an A.P. The last term in this series is found out by dividing...
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(i) First \[1000\] positive integers are: \[1,\text{ }2,\text{ }3,\text{ }4,\text{ }\ldots \ldots ..,\text{ }1000\] This is an A.P with first term a = \[1\] and common difference d = \[1\] We know...
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We know that, \[{{S}_{n}}~=\text{ }n/2\text{ }\times \text{ }\left[ 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right]\] So, \[\begin{array}{*{35}{l}} {{S}_{10}}~=\text{ }10/2\text{...
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Given, \[{{S}_{n}}~=\text{ }6n\text{ }\text{ }{{n}^{2}}\] Now, \[{{S}_{1}}~=\text{ }6\left( 1 \right)\text{ }\text{ }{{\left( 1 \right)}^{2}}~=\text{ }6\text{ }\text{ }1\text{ }=\text{ }5\] So, the...
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Given, \[{{S}_{6}}~=\text{ }42\text{ }and\text{ }{{T}_{10}}/{{T}_{30}}~=\text{ }1/3\] We know that, \[\begin{array}{*{35}{l}} {{S}_{n}}~=\text{ }\left( n/2 \right)\text{ }\left( 2a\text{ }+\text{...
![\[{{\mathbf{a}}_{\mathbf{1}}},\text{ }{{\mathbf{a}}_{\mathbf{2}}},\text{ }{{\mathbf{a}}_{\mathbf{3}}},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b77d0d4cc438a693ae76017b90fc5f9a_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, nth term is \[3+4n\] So, \[{{\mathbf{a}}_{\mathbf{n}}}~=\text{ }\mathbf{3}\text{ }+\text{ }\mathbf{4n}\] Now, we start giving values, \[1,2,3.....\] in the place...
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From the question it is given that, \[\begin{array}{*{35}{l}} {{S}_{6}}~=\text{ }36 \\ {{S}_{16}}~=\text{ }256 \\ \end{array}\] We know that, \[\begin{array}{*{35}{l}} {{S}_{n}}~=\text{ }\left(...
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From the question it is given that, \[{{T}_{2}}~=\text{ }14,\text{ }{{T}_{3}}~=\text{ }18\] So, common difference d = \[{{T}_{3}}~\text{ }{{T}_{2}}\] \[\begin{array}{*{35}{l}} =\text{ }18\text{...
From the question it is given that, \[{{T}_{4}}~=\text{ }8\text{ }and\text{ }{{T}_{6}}~=\text{ }14\] ⇒ \[a\text{ }+\text{ }3d\text{ }=\text{ }8\]… (i) ⇒ \[a\text{ }+\text{ }5d\text{ }=\text{ }14\]…...
From the question it is given that, \[{{T}_{4}}~=\text{ }8\text{ }and\text{ }{{T}_{6}}~=\text{ }14\] ⇒ \[a\text{ }+\text{ }3d\text{ }=\text{ }8\]… (i) ⇒ \[a\text{ }+\text{ }5d\text{ }=\text{ }14\]…...
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From the question it is given that, Common difference d = \[7\] \[\begin{array}{*{35}{l}} {{a}_{22}}~=\text{ }149 \\ n\text{ }=\text{ }22 \\ \end{array}\] we know that, \[\begin{array}{*{35}{l}}...
![\[\mathbf{25},\text{ }\mathbf{22},\text{ }\mathbf{19},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-335b0c7e4f6bb18f7753d6d224e01478_l3.png "Rendered by QuickLaTeX.com")
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![\[\mathbf{24},\text{ }\mathbf{21},\text{ }\mathbf{18},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-11fb115af5a799a8854997a7b11f3f57_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, First term a = \[25\] Common difference d = \[22\text{ }\text{ }25\text{ }=\text{ }\text{ }3\] Sum = \[116\] \[\begin{array}{*{35}{l}} {{S}_{n}}~=\text{ }\left(...
![\[\mathbf{x}\text{ }:\text{ }\mathbf{1}\text{ }+\text{ }\mathbf{4}\text{ }+\text{ }\mathbf{7}\text{ }+\text{ }\mathbf{10}\text{ }+\text{ }\ldots \text{ }+\text{ }\mathbf{x}\text{ }=\text{ }\mathbf{287}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-74233c773a531fa9ebf3f17116d38f65_l3.png "Rendered by QuickLaTeX.com")
From the question, First term a = \[1\] Difference d = \[4\text{ }\text{ }1\text{ }=\text{ }3\] n = x \[\begin{array}{*{35}{l}} x\text{ }=\text{ }a\text{ }=\text{ }\left( n\text{ }\text{ }1...
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From the question it is give that, First term a = \[17\] Last term (l) = \[350\] Common difference d = \[9\] We know that, l = \[{{T}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1...
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From the question it is give that, First term a = \[5\] Last term = \[45\] Then, sum = \[400\] We know that, last term = a + (n – 1)d \[\begin{array}{*{35}{l}} 45\text{ }=\text{ }5\text{ }+\text{...
![\[\mathbf{a}\text{ }=\text{ }\mathbf{3},\text{ }\mathbf{n}\text{ }=\text{ }\mathbf{8},\text{ }\mathbf{S}\text{ }=\text{ }\mathbf{192}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d8c5fba8eb3c1721b0df32f5039a86cb_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[3\] n = \[8\] S = \[192\] We know that, \[{{S}_{n}}\] = (n/2) (2a + (n – 1)d) Therefore, common difference d is \[6\].
![\[{{\mathbf{S}}_{\mathbf{9}}}~=\text{ }\mathbf{75}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bc581b3a098f33349b1bd4d2e2f247ad_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, Common difference d = \[5\] \[{{\mathbf{S}}_{\mathbf{9}}}~=\text{ }\mathbf{75}\] We know that, an = a + (n – 1)d \[\begin{array}{*{35}{l}} {{a}_{9}}~=\text{...
![\[\mathbf{a}\text{ }=\text{ }\mathbf{5},\text{ }\mathbf{d}\text{ }=\text{ }\mathbf{3},\text{ }{{\mathbf{a}}_{\mathbf{n}}}~=\text{ }\mathbf{50},\text{ }\mathbf{find}\text{ }\mathbf{n}\text{ }\mathbf{and}\text{ }{{\mathbf{S}}_{\mathbf{n}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4dbcc6ecd7c0107506e2499f0e6be586_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{a}\text{ }=\text{ }\mathbf{7},\text{ }{{\mathbf{a}}_{\mathbf{13}}}~=\text{ }\mathbf{35}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-79838ecc8b39338c868bac778f522d3c_l3.png "Rendered by QuickLaTeX.com")
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From the question, First term a = \[5\] Then common difference d = \[3\] \[{{a}_{n}}~=\text{ }50\], We know that, \[{{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d\]...
![\[\mathbf{34}\text{ }+\text{ }\mathbf{32}\text{ }+\text{ }\mathbf{30}\text{ }+\text{ }\ldots \text{ }+\text{ }\mathbf{10}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cd353cc5d51117188766d56e7a2ea462_l3.png "Rendered by QuickLaTeX.com")
![\[\text{ }\mathbf{5}\text{ }+\text{ }\left( \text{ }\text{ }\mathbf{8} \right)\text{ }+\text{ }\left( \text{ }\text{ }\mathbf{11} \right)\text{ }+\text{ }\ldots \text{ }+\text{ }\left( \text{ }\text{ }\mathbf{230} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c93762108fa04b23aa6771ff8df4cd01_l3.png "Rendered by QuickLaTeX.com")
From the question, First term a = \[34\], Difference d = \[32\text{ }\text{ }34\text{ }=\text{ }-2\] So, common difference d = \[-2\] Last term Tn = 10 We know that, \[{{T}_{n}}\] = a + (n – 1)d...
![\[\mathbf{2},\text{ }\mathbf{7},\text{ }\mathbf{12},\text{ }\ldots \text{ }\mathbf{to}\text{ }\mathbf{10}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-358369381f93e9eae617407d4a040d07_l3.png "Rendered by QuickLaTeX.com")
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From the question, First term a = \[2\] Then, d = \[7\text{ }\text{ }2\text{ }=\text{ }5\] \[12\text{ }\text{ }7\text{ }=\text{ }5\] So, common difference d = \[5\] \[\begin{array}{*{35}{l}} ...
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From the question it is given that, sum of the first three terms of an A.P. is \[33\]. Let us assume the \[3\] numbers which are in A.P. are, a – d, a, a + d Now adding \[3\] numbers = a – d + a + a...
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From the question it is given that, sum of three numbers in A.P. = \[30\] The ratio of first number to the third number is \[3:7\] Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d...
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From the question it is given that, The sum of three numbers in A.P. = \[3\] Given, Their product = \[-35\] Let us assume the \[3\] numbers which are in A.P. are, a – d, a, a + d Now adding \[3\]...
![\[\mathbf{n}\text{ }\text{ }\mathbf{2},\text{ }\mathbf{4n}\text{ }\text{ }\mathbf{1}\text{ }\mathbf{and}\text{ }\mathbf{5n}\text{ }+\text{ }\mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ab4cbc685bed7c2dca72739baad1c169_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, \[\mathbf{n}\text{ }\text{ }\mathbf{2},\text{ }\mathbf{4n}\text{ }\text{ }\mathbf{1}\text{ }\mathbf{and}\text{ }\mathbf{5n}\text{ }+\text{ }\mathbf{2}\] are in...
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The two digits numbers divisible by \[3\] are, \[12,\text{ }15,\text{ }18,\text{ }21,\text{ }24,\ldots ..,99\]. The above numbers are A.P. So, first number a = \[12\] Common difference d =...
![\[\mathbf{3},\text{ }\mathbf{10},\text{ }\mathbf{17},\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-484d8b5174c4bd394b7791fab2cb16a2_l3.png "Rendered by QuickLaTeX.com")
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From the question it is given that, First term a = \[3\] Common difference d = \[10\text{ }\text{ }3\text{ }=\text{ }7\] Then, \[\begin{array}{*{35}{l}} {{T}_{13}}~=\text{ }a\text{ }+\text{ }12d \\...
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Froom the question it is given that, \[{{T}_{8}}~=\text{ }0\] We have to prove that, \[{{38}^{th}}\] term is triple of its \[{{18}^{th}}\] term = \[{{T}_{38}}~=\text{ }3{{T}_{18}}\]...
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From the question it I s given that, \[{{T}_{10}}~=\text{ }41\] \[{{T}_{10}}~=\text{ }a\text{ }+\text{ }9d\text{ }=\text{ }41\]… [equation (i)] \[\begin{array}{*{35}{l}} {{T}_{15}}~=\text{ }a\text{...
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From the question it is given that, \[\begin{array}{*{35}{l}} {{T}_{9}}~=\text{ }1/7 \\ {{T}_{7}}~=\text{ }1/9 \\ \end{array}\] Let us assume ‘a’ be the first term and ‘d’ be the common...
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From the question it is given that, \[\begin{array}{*{35}{l}} {{T}_{11}}~=\text{ }38 \\ {{T}_{6}}~=\text{ }73 \\ \end{array}\] Let us assume ‘a’ be the first term and ‘d’ be the common difference,...
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From the question it is given that, First term a = \[12\] \[{{\mathbf{7}}^{\mathbf{th}}}\] term is 24 less than the \[{{11}^{th}}\] term = \[{{T}_{11}}~\text{ }{{T}_{7}}~=\text{ }24\]...
From the question it is given that, \[{{T}_{3}}~=\text{ }16\] The 7th term exceeds the 5th term by \[12\text{ }=\text{ }{{T}_{7}}~\text{ }{{T}_{5}}~=\text{ }12\] We know that, \[{{T}_{n}}~=\text{...
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From the question, The first term a = \[53\] Then, difference d \[\begin{array}{*{35}{l}} =\text{ }48\text{ }\text{ }53\text{ }=\text{ }-5 \\ =\text{ }43\text{ }\text{ }48\text{ }=\text{ }-5 \\...
![\[-\mathbf{4}/\mathbf{3},\text{ }-\mathbf{1},\text{ }-\mathbf{2}/\mathbf{3},\text{ }\ldots ,4\frac{1}{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ecd4ef61b00e22b96a69fd4e79726c98_l3.png "Rendered by QuickLaTeX.com")
From the question, Last term (nth) = \[4\frac{1}{3}\] = \[13/3\] First term a = \[-4/3\] Then, difference \[\begin{array}{*{35}{l}} d\text{ }=\text{ }-1\text{ }\text{ }\left( -4/3 \right)\text{...
![\[\mathbf{3},\text{ }\mathbf{8},\text{ }\mathbf{13},\text{ }\ldots ,\text{ }\mathbf{253}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a55e050f060fc9cc02b5e72b900c8913_l3.png "Rendered by QuickLaTeX.com")
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Let us assume \[253\text{ }as\text{ }{{n}^{th}}~\] term. From the question, The first term a = \[3\] Then, difference d \[\begin{array}{*{35}{l}} =\text{ }8\text{ }\text{ }3\text{ }=\text{ }5 \\...
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![\[\mathbf{11},\text{ }\mathbf{8},\text{ }\mathbf{5},\text{ }\mathbf{2},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dc20ed1283e7a15fd1f94c4ca71dbd94_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{7},\text{ }\mathbf{10},\text{ }\mathbf{13},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-55ad5f443ef2928798b6d896554e9844_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, The first term a = \[11\] Then, difference d = \[8\text{ }\text{ }11\text{ }=\text{ }-3\] \[\begin{array}{*{35}{l}} 5\text{ }\text{ }8\text{ }=\text{ }-3 \\...
![\[\mathbf{3},\text{ }\mathbf{8},\text{ }\mathbf{13},\text{ }\mathbf{18},\text{ }\ldots \text{ }\mathbf{is}\text{ }\mathbf{78}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-80e13e80ac0c09644fc3da63a7d68d9e_l3.png "Rendered by QuickLaTeX.com")
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Let us assume \[78\text{ }as\text{ }{{n}^{th}}~\] term. From the question, The first term a = \[3\] Then, difference d \[\begin{array}{*{35}{l}} ~=\text{ }8\text{ }\text{ }3\text{ }=\text{ }5 \\...
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From the question it is given that, The \[\mathbf{1}{{\mathbf{8}}^{\mathbf{th}}}\] term = \[-5\] Then, common difference d = \[-3\] \[\begin{array}{*{35}{l}} {{T}_{n}}~=\text{ }a\text{ }+\text{...
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From the question, The first term a = \[5\] Then, difference d = \[2\text{ }\text{ }5\text{ }=\text{ }\text{ }3\] \[\begin{array}{*{35}{l}} -1\text{ }\text{ }3\text{ }=\text{ }-3 \\ \text{ }4\text{...
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From the question, The first term a = \[1\] Then, difference d = \[6\text{ }\text{ }1\text{ }=\text{ }5\] \[\begin{array}{*{35}{l}} 11\text{ }\text{ }6\text{ }=\text{ }5 \\ 16\text{ }\text{...
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From the question it is given that, nth term is \[\mathbf{7}\text{ }\text{ }\mathbf{3K}\] So, \[{{T}_{n}}~=\text{ }7\text{ }\text{ }3n\] Now, we start giving values, \[1,\text{ }2,\text{ }3,\text{...
![\[\text{ }\mathbf{10},\text{ }-\mathbf{6},\text{ }-\mathbf{2},\text{ }\mathbf{2},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a47f30b42de5e76e5094b8a512c0da62_l3.png "Rendered by QuickLaTeX.com")
![\[{{\mathbf{1}}^{\mathbf{2}}},\text{ }{{\mathbf{3}}^{\mathbf{2}}},\text{ }{{\mathbf{5}}^{\mathbf{2}}},\text{ }{{\mathbf{7}}^{\mathbf{2}}},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fa7773dff892c9c24dcef3e75defe62f_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[-10\] Then, difference d = \[-6\text{ }\text{ }\left( -\text{ }10 \right)\text{ }=\text{ }\text{ }6\text{ }+\text{ }10\text{ }=\text{ }4\]...
![\[~\mathbf{2},\text{ }\mathbf{4},\text{ }\mathbf{8},\text{ }\mathbf{16},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-63b3f78a0d900a4b23c2714eacf2f7b7_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{2},\text{ }\mathbf{5}/\mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{7}/\mathbf{2},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-57010b1dbcdebcc341c1bf1e7d02e0b1_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[2\] Then, difference d = \[4\text{ }\text{ }2\text{ }=\text{ }2\] \[\begin{array}{*{35}{l}} 8\text{ }\text{ }4\text{ }=\text{ }4 \\ 16\text{...
![\[\mathbf{4},\text{ }\mathbf{10},\text{ }\mathbf{16},\text{ }\mathbf{22},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eeb2dfcc5c6fcdaf2fd8fec574cc43c0_l3.png "Rendered by QuickLaTeX.com")
![\[-\mathbf{2},\text{ }\mathbf{2},\text{ }-\mathbf{2},\text{ }\mathbf{2},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5a693055e210164707bc3bc0fb3fd45_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[4\] Then, difference d = \[10\text{ }\text{ }4\text{ }=\text{ }6\] \[\begin{array}{*{35}{l}} 16\text{ }\text{ }10\text{ }=\text{ }6 \\ 22\text{...
![\[\mathbf{a}\text{ }=\text{ }\mathbf{4},\text{ }\mathbf{d}\text{ }=\text{ }-\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cd11428ce5536ceabe7ba182b59f4793_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{a}\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2},\text{ }\mathbf{d}\text{ }=\text{ }-\mathbf{1}/\mathbf{6}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-748d5828a9ae963d9236004a454a172b_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[4\] Common difference d = \[-3\] Then the first four terms are = \[4\text{ }+\text{ }\left( -3 \right)\text{ }=\text{ }4\text{ }\text{ }3\text{...
![\[\mathbf{a}\text{ }=\text{ }\mathbf{10},\text{ }\mathbf{d}\text{ }=\text{ }\mathbf{10}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-abedf1e2655a008f845d456dcd8488d0_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{a}\text{ }=\text{ }-\mathbf{2},\text{ }\mathbf{d}\text{ }=\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-af80e0707721692c501f9ffbe0f48b8f_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that, First term a = \[10\] Common difference d = \[10\] Then the first four terms are = \[10+10=20\] \[\begin{array}{*{35}{l}} 20\text{ }+\text{ }10\text{ }=\text{...
![\[-\mathbf{3}.\mathbf{2},\text{ }-\mathbf{3},\text{ }-\mathbf{2}.\mathbf{8},\text{ }-\mathbf{2}.\mathbf{6},\text{ }\ldots .\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2dc77beaac096a22ecc800de65d5cb39_l3.png "Rendered by QuickLaTeX.com")
From the question, The first term a = \[-3.2\] Then, difference d = -3 – (-3.2) = -3 + 3.2 = 0.2 \[\begin{array}{*{35}{l}} -2.8\text{ }\text{ }\left( -3 \right)\text{ }=\text{ }-2.8\text{ }+\text{...
![\[\mathbf{3},\text{ }\mathbf{1},\text{ }\text{ }\mathbf{1},\text{ }\text{ }\mathbf{3},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5930c7a6164b8526581bcf0a3cb35a52_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{1}/\mathbf{3},\text{ }\mathbf{5}/\mathbf{3},\text{ }\mathbf{9}/\mathbf{3},\text{ }\mathbf{13}/\mathbf{3},\text{ }\ldots .\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a7e7c7cd15b13f4a3d5375009d35102f_l3.png "Rendered by QuickLaTeX.com")
From the question, The first term a = \[3\] Then, difference d = \[1\text{ }\text{ }3\text{ }=\text{ }\text{ }2\] \[\begin{array}{*{35}{l}} \text{ }1\text{ }\text{ }1\text{ }=\text{ }\text{ }2 \\...
Solution: The acetylation of —NH2 group of aniline reduces its activating effect because the lone pair of electrons on the nitrogen of acetanilide interacts with oxygen atom due to resonance.
Solution: Hinsberg's reagent is benzene sulphonyl chloride, also known as $C_6H_5SOCl$. To distinguish between primary, secondary, and tertiary amines, Hinsberg's reagent is used.
Solution: The main product is $C_6H_5CH_2OH$ When $C_6H_5CH_2NH_2$ reacts with HNO2, it produces an unstable diazonium salt, which is then converted to alcohol. When $C_6H_5CH_2NH_2$ reacts with...
Solution: The nitration of organic compounds is done with a nitration mixture, which is a 1:1 solution of HNO3 and H2SO4. In the nitration of benzene, it acts as a base and provides electrophile.
The correct answer is (b)
, subject to the constraints 
and 
The feasible region determined by $x+3 y \geq 6, x-3 y \leq 3,3 x+4 y \leq 24$ $-3 x+2 y \leq 6,5 x+y \geq 5, x \geq 0$ and $y \geq 0$ is given by The corner points of the feasible region are $A(4 /...
Marks obtained 0-10 10-20 20-30 30-40 40-50 No. of students 8 10 22 40 20 Hence determine: (i) the median (ii) the pass marks if 85% of the students pass. (iii) the marks which 45% of the...
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 3 8 12 14 10 6 5 2 Also state the median class Solution: Arranging the data in cumulative frequency table. Marks Frequency Cumulative...
Daily wages in Rs 0-10 10-20 20-30 30-40 40-50 50-60 No. of employees 1 8 10 5 4 2 Estimate the modal daily wages for this distribution by a graphical method. Solution: Daily wages in Rs. No. of...
Age in years 12 13 14 15 16 17 18 No. of students 2 3 5 6 4 3 2 Solution: Age in years xi No. of students fi Cumulative frequency fixi 12 2 2 24 13 3 5 39 14 5 10 70 15 6 16 90 16 4 20 64 17 3 23 51...
Solution: Here n = 9 which is odd. Median = ((n+1)/2)th term Median = ((9+1)/2)th term Median = (10/2)th term Median = 5 th term Median = 5 Mode is the number which appears most often in a set of...
Solution: Arranging data in ascending order 1,3,5,6,8,9,10,12,13,15,17,18,20,21,21,23 Here n = 16 which is even (i) So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (16/2 th term +...
Solution: Arranging numbers in ascending order 17,26,29,32,33,34,45,56,60 Here n = 9 which is odd. So Median =( (n+1)/2) th term Median = ((9+1)/2)th term Median = (10/2)th term Median = 5th term...
Solution: Arranging numbers in ascending order 3,4,5,x,8,9,11 Here n = 7 which is odd Given median = 6 So Median =( (n+1)/2) th term 6 = ((7+1)/2)th term 6 = ((8/2)th term 6 = 4th term 6 = x Hence...
Given $x-y \leq 3$ $\Rightarrow-y \leq 3-x$ Multiplying by minus on both the sides, we'll get $\begin{array}{l} y \geq-3+x \\ y \geq x-3 \end{array}$ Consider the equation $y=x-3$. Finding points on...
Solution: Here n = 10, which is even Median = 25 So Median = ½ ( n/2 th term + ((n/2)+1)th term) 25 = ½ (( 10/2 )th term + (10/2)+1)th term) 25 = ½ (( 5 )th term + (6)th term) 25 = ½ (x+2 + x+4) 25...
Diameter in mm 32-36 37-41 42-46 47-51 52-56 57-61 62-66 No. of screws 15 17 p 25 q 20 30 Solution: Given mean = 51.2 mm The given distribution is not continuous. Adjustment factor = (37-36)/2 = ½ =...
Expenditure in Rs 140-160 160-180 180-200 200-220 220-240 No. of families 5 25 f1 f2 5 Solution: Given mean = 188 Class Frequency fi Class mark xi fixi 140-160 5 150 750 160-180 25 170 4250 180-200...
Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 5 8 p 12 7 8 Solution: Class Frequency fi Class mark xi fixi 0-20 5 10 50 20-40 8 30 240 40-60 p 50 50p 60-80 12 70 840 80-100 7 90 630 100-120...
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Students 2 4 5 16 20 10 6 8 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Marks Students fi Class mark xi fixi...
Age in years 25-29 30-34 35-39 40-44 45-49 50-54 55-59 No. of persons 4 14 22 16 6 5 3 Solution: The given distribution is not continuous. Adjustment factor = (30-29)/2 = ½ = 0.5 We subtract 0.5...
Variate (xi) 13 15 17 19 20+p 23 Frequency (fi) 8 2 3 4 5p 6 Solution: Variate (xi) Frequency (fi) fi xi 13 8 104 15 2 30 17 3 51 19 4 76 20+p 5p 5p2+100p 23 6 138 Total Ʃfi = 23+5p Ʃfi xi =...
Variate (xi) 10 15 20 25 35 Frequency (fi) 3 10 p 7 5 Solution: Variate (xi) Frequency (fi) fx 10 3 30 15 10 150 20 p 20p 25 7 175 35 5 175 Total Ʃfi = 25+p Ʃfi xi = 530+20p Mean = Ʃfx/Ʃf 20.6 =...
Height (in cm) 65 66 67 68 69 70 71 72 73 No. of children 1 4 5 7 11 10 6 4 2 Calculate the mean height for this distribution correct to one place of decimal. Solution: Height x No. of children f...
No. of matches 41 42 43 44 45 46 No. of boxes 5 8 13 12 7 5 Calculate the mean number of matches per box. Solution: No. o f matches x No. of boxes f fx 41 5 205 42 8 336 43 13 559 44 12 528 45 7 315...
Solution: Total number of students = 50 No. of boys = 40 No. of girls = 50-40 = 10 Average weight of 50 students = 44 kg So sum of weight = 44×50 = 2200 kg Average weight of girls = 40 kg So sum of...
Solution: Average height of 30 students = 150 cm So sum of height = 150×30 = 4500 Difference between correct value and wrong value = 165-135 = 30 So actual sum = 4500+30 = 4530 So actual mean =...
Solution: Given the mean of 20 numbers = 18 Sum of numbers = 18×20 = 360 If 3 is added to each of first 10 numbers, then new sum = (3×10)+360 = 30+360 = 390 New mean = 390/20 = 19.5 Hence the mean...
Solution: Marks scored in English = 36 Marks scored in Civics = 44 Marks scored in Mathematics = 75 Marks scored in Science = x No. of subjects = 4 Average marks = sum of marks / No. of subjects =...
Height in cm 121-130 131-140 141-150 151-160 161-170 171-180 No. of pupils 12 16 30 20 14 8 Draw the ogive for the above data and from it determine the median (use graph paper). Solution: We write...
Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180 No. of students 12 20 30 38 24 16 12 8 Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2...
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 5 9 16 22 26 18 11 6 4 3 Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for ogive...
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 3 7 12 17 23 14 9 6 5 4 Draw an ogive on a graph sheet and from it determine the : (i) median (ii) lower quartile ...
Weight 40-45 45-50 50-55 55-60 60-65 65-70 70-75 75-80 Frequency 5 17 22 45 51 31 20 9 Use your ogive to estimate the following: (i) The percentage of students weighing 55 kg or more. (ii) The...
Monthly income No. of employees 6000-7000 20 7000-8000 45 8000-9000 65 9000-10000 95 10000-11000 60 11000-12000 30 12000-13000 5 Draw an ogive of the given distribution on a graph sheet taking 2 cm...
marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 5 11 10 20 28 37 40 29 14 6 Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm =...
Wages in Rs 400-450 450-500 500-550 550-600 600-650 650-700 700-750 No. of workers 2 6 12 18 24 13 5 Use a graph paper to draw an ogive for the above distribution. ( a scale of 2 cm = Rs 50 on x-...
Scores 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of shooters 9 13 20 26 30 22 15 10 8 7 Use your graph to estimate the following: (i) The median. (ii) The interquartile...
Weight in kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120 No. of workers 4 7 11 14 6 5 3 Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis , and 2 cm = 5...
Age( yrs) 5-15 15-25 25-35 35-45 45-55 55-65 65-75 No. of casualities due to accidents 6 10 15 13 24 8 7 (i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10...
Weight (gm) 50-60 60-70 70-80 80-90 90-100 100-110 110-120 120-130 Frequency 8 10 12 16 18 14 12 10 (i) Calculate the cumulative frequencies. (ii) Draw the cumulative frequency curve and from it...
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 No. of students 3 8 12 14 10 6 5 2 Solution: We write the given data in cumulative frequency table. Marks No of students f Cumulative frequency...
Height ( in cm ) 150-155 155-160 160-165 165-170 170-175 175-180 180-185 No. of workers 6 12 18 20 13 8 6 (i) Determine the cumulative frequencies. (ii) Draw the cumulative frequency curve on a...
Marks obtained 24-29 29-34 34-39 39-44 44-49 49-54 54-59 No. of students 1 2 5 6 4 3 2 Solution: We write the given data in cumulative frequency table. Marks obtained No of students Cumulative...
Class intervals 1-10 11-20 21-30 31-40 41-50 51-60 Frequency 3 5 8 7 6 2 Solution: The given distribution is not continuous. Adjustment factor = (11-10)/2 = ½ = 0.5 We subtract 0.5 from lower limit...
Height ( in cm ) 150-160 160-170 170-180 180-190 190-200 No. of students 8 3 4 10 2 Solution: We write the given data in cumulative frequency table. Height in cm No of students Cumulative frequency...
Mid value 12 18 24 30 36 42 48 Frequency 20 12 8 24 16 8 12 Also state the modal class. Solution: Mid value Frequency 12 20 18 12 24 8 30 24 36 16 42 8 48 12 Here mid value and frequency is given....
Wt. in kg 40-44 45-49 50-54 55-59 60-64 65-69 No. of students 2 8 12 10 6 4 Hence estimate the modal weight. Solution: The given distribution is not continuous. Adjustment factor = (45-44)/2 = ½ =...
Pocket expenses (in Rs) Number of students (Frequency ) 0-5 10 5-10 14 10-15 28 15-20 42 20-25 50 25-30 30 30-35 14 35-40 12 Draw a histogram representing the above distribution and estimate the...
IQ score 80-90 90-100 100-110 110-120 120-130 130-140 No. of students 6 9 16 13 4 2 Draw a histogram for the above data and estimate the mode. Solution: Construct histogram using given data....
Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 8 10 5 4 3 Solution: Construct histogram using given data. Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 8 10 5 4 3 Represent classes...
Marks 50-60 60-70 70-80 80-90 90-100 No. of students 4 8 14 19 5 Draw a histogram for the above data using a graph paper and locate the mode. (2011) Solution: Construct histogram using given data....
Height (in cm) 140-150 150-160 160-170 170-180 180-190 No. of students 7 6 4 10 2 Solution: Construct histogram using given data. Height (in cm) 140-150 150-160 160-170 170-180 180-190 No. of...
Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5 Solution: Construct histogram using given data. Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5 Represent class on...
Weekly wages (in Rs) 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 No. of workers 5 20 10 10 9 6 12 8 Calculate: (i) The mean. (ii) the modal class (iii) the number of workers getting weekly wages...
Class interval 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 Frequency 5 20 10 10 9 6 12 8 Solution: (i) Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 67.5 Class size (h)...
Score 0 1 2 3 4 5 No. of shots 0 3 6 4 7 5 (i) What was his modal score ? (ii) What was his median score ? (iii) What was his total score ? (iv) What was his mean ? Solution: We write the marks in...
Marks obtained 5 6 7 8 9 10 No. of students 3 9 6 4 2 1 Solution: We write the marks in cumulative frequency table. Marks x Frequency (f) fx Cumulative frequency 5 3 15 3 6 9 54 12 7 6 42 18 8 4 32...
Marks 0 1 2 3 4 5 No. of students 1 3 6 10 5 5 Calculate the mean, median and mode of the above distribution. Solution: We write the data in cumulative frequency table. Marks x Frequency (f)...
x 10 11 12 13 14 15 f 1 4 7 5 9 3 Solution: We write the data in cumulative frequency table. x Frequency (f) Cumulative frequency 10 1 1 11 4 5 12 7 12 13 5 17 14 9 26 15 3 29 Here number of...
Solution: Given data is 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8 Number of observations, n = 16 Mean = Ʃxi/n = (0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8)/16 = 64/16 = 4 Hence the mean is 4. Here n = 16...
Solution: (i)We arrange given marks in ascending order 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19 16 appears maximum number of times. Hence his modal mark is 16. (ii)Here number of observations, n =...
Solution: Given data in ascending order: 13, 35, 43, 46, x, x +4, 55, 61,71, 80 Given median = 48 Number of observations, n = 10 which is even. median = ½ ( n/2 th term + ((n/2)+1)th term) 48 = ½...
Solution: We arrange given data in ascending order 1, 2, 3, 3, 3, 4, 5, 5, 6, 7 Mean = Ʃxi/n = (1+2+3+3+3+4+5+5+6+7)/10 = 39/10 = 3.9 Here number of observations, n = 10 which is even. So median = ½...
Solution: We arrange given data in ascending order 6, 6, 7, 8, 10, 10, 10, 11, 13 Mean = Ʃxi/n = (6+6+7+8+10+10+10+11+13)/9 = 81/9 = 9 Hence the mean is 9. Here number of observations, n = 9 which...
Solution: We arrange given data in ascending order 1, 1, 2, 2, 3, 3, 3, 6 Mean = Ʃxi/n = (1+1+2+2+3+3+3+6)/8 = 21/8 = 2.625 Hence the mean is 2.625. Here number of observations, n = 8 which is even....
Solution: Mode is the number which appears most often in a set of numbers. (i)Given set is 3, 2, 0, 1, 2, 3, 5, 3. In this set, 3 occurs maximum number of times. Hence the mode is 3. (ii) Given set...
Variate 25 31 34 40 45 48 50 60 Frequency 3 8 10 15 10 9 6 2 Solution: We write the variates in cumulative frequency table. Variate Frequency (f) Cumulative frequency 25 3 3 31 8 11 34 10 21 40 15...
Variate 15 18 20 22 25 27 30 Frequency 4 6 8 9 7 8 6 Solution: We write the variates in cumulative frequency table. Variate Frequency (f) Cumulative frequency 15 4 4 18 6 10 20 8 18 22 9 27 25 7 34...
Solution: Arranging the observations in ascending order 21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53 Here n = 19 which is odd. (i)Median = ((n+1)/2)th term = (19+1)/2 =...
Number 5 10 15 20 25 30 35 Frequency 1 2 5 6 3 2 1 Solution: We write the numbers in cumulative frequency table. Marks (x) No. of students (f) Cumulative frequency fx 5 1 1 5 10 2 3 20 15 5 8 75 20...
Marks 20 70 50 60 75 90 40 No. of students 8 12 18 6 9 5 12 Calculate the median marks. Solution: We write the marks in ascending order in cumulative frequency table. Marks No. of students (f)...
Marks 35 45 50 64 70 72 No. of students 3 5 8 10 5 5 Solution: We write the distribution in cumulative frequency table. Marks No. of students (f) Cumulative frequency 35 3 3 45 5 8 50 8 16 64 10 26...
Wages per day in Rs. 38 45 48 55 62 65 No. of workers 14 8 7 10 6 2 Solution: We write the distribution in cumulative frequency table. Wages per day in Rs. No. of workers (f) Cumulative frequency 38...
Solution: (i) Mean of 1, 7, 5, 3, 4, 4 is m. Here n = 6 Mean, m = (1+7+5+3+4+4)/6 m = 24/6 m = 4 Given the numbers 3, 2, 4, 2, 3, 3, p have mean m-1. So m-1 = (3+2+4+2+3+3+p)/7 4-1 = (17+p)/7 3 =...
Solution: Observation are as follows : 11, 12, 14, (x-2), (x+4), (x+9), 32, 38, 47 n = 9 Here n is odd. So median = ((n+1)/2)th term = (9+1)/2 )th term = 5th term = x+4 Given median = 24 x+4 = 24 x...