ML Aggarwal

(i) Find the sum of first

    \[51\]

terms of the A.P. whose second and third terms are

    \[14\]

and

    \[18\]

, respectively. (ii) The

    \[{{4}^{th}}\]

term of A.P is

    \[22\]

and

    \[{{15}^{th}}\]

term is

    \[66\]

. Find the first term and the common difference. Hence, find the sum of first 8 term of the A.P.

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{...

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(i) The first term of an A.P. is 5, the last term is

    \[45\]

and the sum is

    \[400\]

. Find the number of terms and the common difference. (ii) The sum of first

    \[15\]

terms of an A.P. is

    \[750\]

and its first term is

    \[15\]

. Find its

    \[20\]

th term.

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{...

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(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{...

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(i)If the common difference of an A.P. is

    \[-3\]

and the

    \[\mathbf{1}{{\mathbf{8}}^{\mathbf{th}}}\]

term is

    \[-5\]

, then find its first term. (ii) If the first term of an A.P. is

    \[-18\]

and its

    \[10\]

th term is zero, then find its common difference.

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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The table shows the distribution of scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis)

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...

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A boy scored the following marks in various class tests during a term each test being marked out of 20: 15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16
(i) What are his modal marks ?
(ii) What are his median marks ?
(iii) What are his mean marks ?

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 =...

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The marks of 10 students of a class in an examination arranged in ascending order are as follows: 13, 35, 43, 46, x, x +4, 55, 61,71, 80 If the median marks is 48, find the value of x. Hence, find the mode of the given data. (2017)

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...

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The daily wages in (rupees of) 19 workers are 41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35. find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range

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 =...

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(a) The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes years. What is the age of the girl ?
(b) In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that marks of one student was wrongly copied as 21 instead of 29. Find the correct mean.

Solution: (a)Given mean age = 13 Number of students = 33 Sum of ages = mean ×number of students = 13×33 = 429 After a girl leaves, the mean of 32 students becomes = 207/16 Now sum of ages =...

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The marks obtained by 15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20 find
(i) the mean of their marks.
(ii) the mean of their marks when the marks of each student are increased by 4.
(iii) the mean of their marks when 2 marks are deducted from the marks of each student.
(iv) the mean of their marks when the marks of each student are doubled.

Solution: (i) Marks obtained by students are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20. Number of students = 15 Mean = sum of observations / number of observations...

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