(i) The \[{{15}^{th}}\] term of an A.P. is \[3\] more than twice its \[{{7}^{th}}\] term. If the \[{{10}^{th}}\] term of the A.P. is \[41\], find its nth term. (ii) The sum of \[{{5}^{th}}\] and \[{{7}^{th}}\] terms of an A.P. is \[52\] and the \[{{10}^{th}}\] term is \[46\]. Find the A.P.

(i) The

${{15}^{th}}$

term of an A.P. is

$3$

more than twice its

${{7}^{th}}$

term. If the

${{10}^{th}}$

term of the A.P. is

$41$

, find its nth term. (ii) The sum of

${{5}^{th}}$

and

${{7}^{th}}$

terms of an A.P. is

$52$

and the

${{10}^{th}}$

term is

$46$

. Find the A.P.

(i) The

${{15}^{th}}$

term of an A.P. is

$3$

more than twice its

${{7}^{th}}$

term. If the

${{10}^{th}}$

term of the A.P. is

$41$

, find its nth term. (ii) The sum of

${{5}^{th}}$

and

${{7}^{th}}$

terms of an A.P. is

$52$

and the

${{10}^{th}}$

term is

$46$

. Find the A.P.

From the question it I s given that,

${{T}_{10}}~=\text{ }41$

${{T}_{10}}~=\text{ }a\text{ }+\text{ }9d\text{ }=\text{ }41$

… [equation (i)]

$\begin{array}{*{35}{l}} {{T}_{15}}~=\text{ }a\text{ }+\text{ }14d\text{ }=\text{ }2{{T}_{7}}~+\text{ }3 \\ =\text{ }a\text{ }+\text{ }14d\text{ }=\text{ }2\left( a\text{ }+\text{ }6d \right)\text{ }+\text{ }3 \\ =\text{ }a\text{ }+\text{ }14d\text{ }=\text{ }2a\text{ }+\text{ }12\text{ }d\text{ }+\text{ }3 \\ -3\text{ }=\text{ }2a\text{ }\text{ }a\text{ }+\text{ }12d\text{ }\text{ }14d \\ \end{array}$

$a\text{ }\text{ }2d\text{ }=\text{ }-3$

… [equation (ii)]

Now, subtracting equation (ii) from (i), we get,

$\begin{array}{*{35}{l}} \left( a\text{ }+\text{ }9d \right)\text{ }\text{ }\left( a\text{ }\text{ }2d \right)\text{ }=\text{ }41\text{ }\text{ }\left( -3 \right) \\ a\text{ }+\text{ }9d\text{ }\text{ }a\text{ }+\text{ }2d\text{ }=\text{ }41\text{ }+\text{ }3 \\ 11d\text{ }=\text{ }44 \\ d\text{ }=\text{ }44/11 \\ d\text{ }=\text{ }4 \\ \end{array}$

Then, substitute the value of d is equation (i) to find a,

$\begin{array}{*{35}{l}} a\text{ }+\text{ }9\left( 4 \right)\text{ }=\text{ }41 \\ a\text{ }+\text{ }36\text{ }=\text{ }41 \\ a\text{ }=\text{ }41\text{ }\text{ }36 \\ a\text{ }=\text{ }5 \\ \end{array}$

Therefore, n^th term =

${{T}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d$

$\begin{array}{*{35}{l}} =\text{ }5\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)4 \\ =\text{ }5\text{ }+\text{ }4n\text{ }\text{ }4 \\ =\text{ }1\text{ }+\text{ }4n \\ \end{array}$

(ii) The sum of

${{5}^{th}}$

and

${{7}^{th}}$

terms of an A.P. is

$52$

and the

${{10}^{th}}$

term is

$46$

. Find the A.P.

Solution:-

From the question it is given that,

a₅ + a₇ = 52

(a + 4d) + (a + 6d) = 52

a + 4d + a + 6d = 52

2a + 10d = 52

Divide both the side by

$2$

we get,

$a\text{ }+\text{ }5d\text{ }=\text{ }26$

… equation (i)

Given,

${{a}_{10}}~=\text{ }a\text{ }+\text{ }9d\text{ }=\text{ }46$

$a\text{ }+\text{ }9d\text{ }=\text{ }46$

… equation (ii)

Now subtracting equation (i) from equation (ii),

$\begin{array}{*{35}{l}} \left( a\text{ }+\text{ }9d \right)\text{ }\text{ }\left( a\text{ }+\text{ }5d \right)\text{ }=\text{ }46\text{ }\text{ }26 \\ a\text{ }+\text{ }9d\text{ }\text{ }a\text{ }\text{ }5d\text{ }=\text{ }20 \\ 4d\text{ }=\text{ }20 \\ d\text{ }=\text{ }20/4 \\ d\text{ }=\text{ }5 \\ \end{array}$

Substitute the value of d in equation (i) to find out a,

$\begin{array}{*{35}{l}} a\text{ }+\text{ }5d\text{ }=\text{ }26 \\ a\text{ }+\text{ }5\left( 5 \right)\text{ }=\text{ }26 \\ a\text{ }+\text{ }25\text{ }=\text{ }26 \\ a\text{ }=\text{ }26\text{ }\text{ }25 \\ a\text{ }=\text{ }1 \\ \end{array}$

Then,

$\begin{array}{*{35}{l}} {{a}_{2}}~=\text{ }a\text{ }+\text{ }d \\ =\text{ }1\text{ }+\text{ }5\text{ }=\text{ }6 \\ {{a}_{3}}~=\text{ }{{a}_{2}}~+\text{ }d \\ =\text{ }6\text{ }+\text{ }5 \\ =\text{ }11 \\ \end{array}$

$\begin{array}{*{35}{l}} {{a}_{4}}~=\text{ }{{a}_{3}}~+\text{ }d \\ =\text{ }11\text{ }+\text{ }5 \\ =\text{ }16 \\ \end{array}$

Therefore,

$1,\text{ }6,\text{ }11,\text{ }16,\ldots$

are A.P.

More Material

A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia.
A Concave, concave
B Convex, convex
C Convex, concave
D Concave, convex

Rate of reaction of any substance depends on
(A) active mass
(B) molecular weight
(C) atomic weight
(D) equivalent weight

Formula of oleum is
(A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(C) $\mathrm{H}_{2} \mathrm{SO}_{3}$
(D) $\mathrm{H}_{2} \mathrm{SO}_{5}$

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

Two cells of emf $\varepsilon_{1}$ and $\varepsilon_{2}$ , internal resistance $r_{1}$ and $r_{2}$ , connected in parallel. The equivalent emf of the combination is- (A) $\frac{\varepsilon_{1} r_{1}+\varepsilon_{1} r_{1}}{r_{1}+r_{2}}$ (B) $\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$ (C) $\sqrt{\varepsilon_{1} \times \varepsilon_{2}}$ (D) $\frac{\varepsilon_{1}+\varepsilon_{2}}{2}$

More Material

A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia.
A Concave, concave
B Convex, convex
C Convex, concave
D Concave, convex

Rate of reaction of any substance depends on
(A) active mass
(B) molecular weight
(C) atomic weight
(D) equivalent weight

Formula of oleum is
(A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(C) $\mathrm{H}_{2} \mathrm{SO}_{3}$
(D) $\mathrm{H}_{2} \mathrm{SO}_{5}$

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

If the points $A(a, 0), B(0, b)$ and $C(1,1)$ are collinear, prove that $\frac{1}{a}+\frac{1}{b}=1$

If $A(-2,0), B(0,4)$ and $C(0, k)$ be three points such that area of a $A B C$ is 4 sq units, find the value of $k$ .

Find the value of $k$ for which the area of a ABC having vertices $A(2,-6), B(5,4)$ and $C(k, 4)$ is 35 sq units.

Find the value of $k$ for which the points $A(1,-1), B(2, k)$ and $C(4,5)$ are collinear.

Find the value of $k$ for which the points $P(5,5), Q(k, 1)$ and $R(11,7)$ are collinear.

Find the value of $k$ for which the points $A(3,-2), B(k, 2)$ and $C(8,8)$ are collinear.

Use determinants to show that the following points are collinear. $P(-2,5), Q(-6,-7)$ and $R(-5,-4)$

Sign up now and study all subjects for free

Class

Syllabus

Question Papers

Social Media

Apps