Chapter 9

Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry · 307 exercises

Problem 54

Find an A.P. in which sum of any number of terms is always three times the squared number of these terms.

4 step solution

Problem 56

Sum of certain consecutive odd positive integers is \(57^{2}-13^{2}\). Find them.

4 step solution

Problem 57

150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped the second day, four more workers dropped the third day and so on. It takes 8 more days to finish the work now. Find the number of days in which the work was completed. \

4 step solution

Problem 58

Along a road lie an odd number of stones placed at intervals of 10 meters. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried the job with one of the end stones by carrying them in succession. In carrying all the stones he covered a distance of \(3 \mathrm{~km}\). Find the number of stones.

3 step solution

Problem 59

Balls are arranged in rows to from an equilateral triangle. The first row consists of one ball, the second row of two balls and so on. If 669 more balls are added then all the ball can be arranged in the shape of a square and each of the sides then contains 8 balls less then each side of the triangle did. Determine the initial numbers of balls.

4 step solution

Problem 60

Certain numbers appear in both arithmetic progressions \(17,21,25, \ldots\) and \(16,21,26, \ldots\) Find the sum of first hundred numbers appearing in both progressions.

4 step solution

Problem 61

Let \(S_{n}\) denote the sum of first \(n\) terms of an A.P. If \(S_{2 n}=3 S_{n}\), then find the ratio \(\frac{S_{3 n}}{S_{n}}\)

8 step solution

Problem 62

If \(S_{n}=Q n^{2}+P n\), where \(S_{n}\) denotes the sum of the first \(n\) terms of an A.P., then show that the third term is \(P+5 Q\).

5 step solution

Problem 63

The first and last term of an A.P. are \(a\) and \(l\) respectively. If \(S\) be the sum of all the term of the A.P., show that the common difference is \(\frac{l^{2}-a^{2}}{2 S-(l+a)}\).

9 step solution

Problem 64

Show that the sum of an A.P. whose first term is \(a\), second term is \(b\) and the last term is \(c\) is equal to \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\)

4 step solution

Problem 65

If the ratio of the sum of \(m\) term and \(n\) terms of an A.P. be \(m^{2}: n^{2}\), prove that the ratio of its \(m\) th and \(n\) th terms will be \(2 m-1: 2 n-1\).

5 step solution

Problem 66

The ratio between the sum of \(n\) term of two A.P.'s is \(3 n+8: 7 n+15\). Find the ratio between their 12 th terms. Also find the ratio of their common difference.

3 step solution

Problem 67

The ratio between the sum of \(n\) term of two A.P.'s is \(7 n+1: 4 n+27\). Find the ratio between their \(n\) th terms.

3 step solution

Problem 68

The sum of the first \(n\) term of two A.P.'s are as \(3 n+5: 5 n-9\). Prove that their 4 th terms are equal.

4 step solution

Problem 69

The \(p\) th term of an A.P. is \(a\) and \(q\) th term is \(b\). Prove that sum of its \((p+q)\) terms is \(\frac{p+q}{2}\left[a+b+\frac{a-b}{p-q}\right.\).

6 step solution

Problem 70

If the sums of \(p, q\) and \(r\) terms of an A.P. be \(a, b\) and \(c\) respectively then prove that \(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0\)

3 step solution

Problem 71

If in an A.P the sum of \(p\) terms is equal to sum of \(q\) terms, then prove that the sum of \(p+q\) terms is zero.

3 step solution

Problem 72

In an A.P., of which \(a\) is the first term, if the sum of the first \(p\) terms is zero, show that the sum of the next \(q\) terms is \(-\frac{a(p+q) q}{p-1}\).

4 step solution

Problem 73

The sum of first \(p\) terms of an A.P. is \(q\) and the sum of the first \(q\) terms is \(p\). Find the sum of the first \(p+q\) terms.

4 step solution

Problem 74

Prove that the sum of the latter half of \(2 n\) terms of any A.P. is one-third the sum of \(3 n\) terms of the same A.P.

5 step solution

Problem 75

The sums of \(n\) terms of three arithmetical progressions are \(S_{1}, S_{2}\) and \(S_{3}\). The first term of each is unity and the common differences are 1,2 and 3 respectively. Prove that \(S_{1}+S_{3}=2 S_{2}\).

3 step solution

Problem 76

The sums of \(n, 2 n, 3 n\) terms of an A.P. are \(S_{1}, S_{2}, S_{3}\) respectively. Prove that \(S_{3}=3\left(S_{2}-S_{1}\right)\).

5 step solution

Problem 78

There are \(n\) A.P.'s whose common difference are \(1,2,3, \ldots n\) respectively, the first term of each being unity. Prove that sum of their \(n\) th terms is \(\frac{1}{2} n\left(n^{2}+1\right)\).

4 step solution

Problem 79

If there be \(m\) A.P.'s beginning with unity whose common differences are \(1,2,3, \ldots m\) respectively, show that the sum of their \(n\) th terms is \(\frac{1}{2} m[m n-m+n+1]\).

4 step solution

Problem 80

The sum of \(n\) terms of \(m\) arithmetical progressions are \(S_{1}, S_{2}, S_{3}, \ldots . S_{m}\). The first term and common differences are \(1,2,3, \ldots, m\) respectively. Prove that \(S_{1}+S_{2}+S_{3}+\ldots+S_{m}=\frac{1}{4} m n(m+1)(n+1)\).

5 step solution

Problem 81

If \(S_{1}, S_{2}, S_{3}, \ldots, S_{m}\) are the sums of \(n\) term of \(m\) A.P.'s whose first terms are \(1,2,3, \ldots, m\) and common differences are \(1,3,5, \ldots ., 2 m-1\) respectively. Show that \(S_{1}+S_{2}+S_{3}+\ldots+S_{m}=\frac{1}{2} m n(m n+1)\).

4 step solution

Problem 82

If the sum of \(m\) terms of an A.P. is equal to sum of either the next \(n\) terms or the next \(p\) terms, prove that \((m+n)\left(\frac{1}{m}-\frac{1}{p}\right)=(m+p)\left(\frac{1}{m}-\frac{1}{n}\right) .\)

4 step solution

Problem 83

Show that in arithmetical progression \(a_{1}, a_{2}, a_{3}, \ldots ., a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-a_{4}^{2}+\ldots-a_{2 k}^{2}=\frac{k}{2 k-1}\left(a_{1}^{2}-a_{2 k}{\underline{\phantom{xx}}}^{2}\right)\)

5 step solution

Problem 84

If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\) be an A.P. of non-zero terms prove that i. \(\frac{1}{a_{1} a_{n}}+\frac{1}{a_{2} a_{n-1}}+\frac{1}{a_{3} a_{n-2}}+\ldots+\frac{1}{a_{n} a_{1}}=\frac{2}{a_{1}+a_{n}}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right)\) ii. \(\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\ldots+\frac{1}{a_{n-1} a_{n}}=\frac{n-1}{a_{1} a_{n}}\).

6 step solution

Problem 85

If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\) are in A.P. where \(a_{i}>0\) for all \(i\), show that \(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots \cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}\)

3 step solution

Problem 86

Let the sequence \(a_{1}, a_{2}, a_{3}, \ldots ., a_{n}\) from an A.P. and let \(a_{1}=0\), prove that \(\frac{a_{3}}{a_{2}}+\frac{a_{4}}{a_{3}}+\frac{a_{5}}{a_{4}}+\ldots+\frac{a_{n}}{a_{n-1}}-a_{2}\left(\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{n-2}}\right)=\frac{a_{n-1}}{a_{2}}+\frac{a_{2}}{a_{n-1}}\)

3 step solution

Problem 87

Find fifth of the ten arithmetic means inserted between 1 and 100 .

2 step solution

Problem 88

If \(a, b, c, d, e, f\) are AM's between 2 and 12, then find the value of \(a+b+c+d+e+f\).

4 step solution

Problem 89

The sum of two numbers is \(2 \frac{1}{6}\). An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Find the number of means inserted.

4 step solution

Problem 90

Between 1 and 31 are inserted \(m\) arithmetic means so that the ratio of the 7 th and \((m-1)\) th means is \(5: 9\). Find the value of \(m\)

4 step solution

Problem 91

Prove that the sum of the \(n\) arithmetic means inserted between two quantities is \(n\) times the single arithmetic mean between them.

6 step solution

Problem 92

For what value of \(n, \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) is the arithmetic mean of \(a\) and \(b ?\)

9 step solution

Problem 93

If \(x, 2 x+2,3 x+3\) are in GP, then find the fourth term.

4 step solution

Problem 94

Find three numbers in G.P. whose sum is 65 and whose product is 3375 .

3 step solution

Problem 95

The product of three numbers in G.P. is 125 and sum of their products taken in pairs is \(87 \frac{1}{2}\). Find them.

6 step solution

Problem 96

If the product of three numbers in G.P. is 216 and sum of the products taken in pairs is 156 , find the numbers.

10 step solution

Problem 97

Three numbers are in G.P. whose sum is 70 . If the extremes be each multiplied by 4 and the mean by 5 , they will be in A.P. Find the numbers.

9 step solution

Problem 98

The sum of three numbers in G.P. is 14 . If the first two terms are each increased by 1 and the third term decreased by 1 , the resulting numbers are in A.P. Find the numbers.

9 step solution

Problem 99

Three numbers whose sum is 15 are in A.P. If \(1,4,19\) be added to them respectively, then they are in G.P. Find the numbers.

5 step solution

Problem 100

In a set of four numbers the first three are in G.P. and the last three in A.P. with common difference \(6 .\) If the first number is the same as the fourth, find the four numbers.

5 step solution

Problem 101

Find four numbers in G.P. whose sum is 85 and product is 4096 .

6 step solution

Problem 102

Does there exist a geometric progression containing 27,8 and 12 as three of its terms? If it exists, how many such progressions are possible?

4 step solution

Problem 103

Show that the numbers \(10,11,12\) cannot be the terms of a single G.P. with common ratio not equal to \(1 .\)

3 step solution

Problem 105

In a G.P. the first, third and fifth terms may be considered as the first, fourth and sixteenth terms of an A.P. Determine the fourth term of the A.P., knowing that its first term is 5 and determine \(T_{1}, T_{3}, T_{5}\) of G.P.

6 step solution

Problem 106

The sum of first ten terms of an A.P. is equal to 155 , and sum of the first two terms of a G.P. is 9 , find these progressions if the first term of A.P. is equal to common ratio of G.P. and the first term of G.P. is equal to common difference of A.P.

4 step solution

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