NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions

Exercise 5.3

Match the APs in Column A with their common differences in Column B.

Column A:
(A₁) 2, −2, −6, −10, …
(A₂) \(a = -18,\; n = 10,\; a_n = 0\)
(A₃) \(a = 0,\; a_{10} = 6\)
(A₄) \(a_2 = 13,\; a_4 = 3\)

Column B:
(B₁) \(\dfrac{2}{3}\), (B₂) −5, (B₃) 4, (B₄) −4, (B₅) 2, (B₆) \(\dfrac{1}{2}\), (B₇) 5

Verify each is an AP and write its next three terms.

(i) \(0,\; \dfrac{1}{4},\; \dfrac{1}{2},\; \dfrac{3}{4},\ldots\)

(ii) \(5,\; \dfrac{14}{3},\; \dfrac{13}{3},\; 4,\ldots\)

(iii) \(\sqrt{3},\; 2\sqrt{3},\; 3\sqrt{3},\ldots\)

(iv) \(a+b,\; (a+1)+b,\; (a+1)+(b+1),\ldots\)

(v) \(a,\; 2a+1,\; 3a+2,\; 4a+3,\ldots\)

Write the first three terms of the AP given \(a\) and \(d\).

(i) \(a=\dfrac{1}{2},\; d=-\dfrac{1}{6}\)

(ii) \(a=-5,\; d=-3\)

(iii) \(a=\sqrt{2},\; d=\dfrac{1}{\sqrt{2}}\)

Find \(a, b, c\) such that \(a,\;7,\;b,\;23,\;c\) are in AP.

Determine the AP whose 5th term is 19 and \(a_{13}-a_{8}=20\).

The 26th, 11th and last terms are \(0,\;3\) and \(-\dfrac{1}{5}\), respectively. Find the common difference and number of terms.

\(a_5+a_7=52\) and \(a_{10}=46\). Find the AP.

Find \(a_{20}\) if \(a_1=12\) and \(a_7\) is 24 less than \(a_{11}\).

If \(a_9=0\), prove \(a_{29}=2a_{19}\).

Is 55 a term of AP \(7,10,13,\ldots\)? If yes, which term?

Determine \(k\) so that \(k^2+4k+8\), \(2k^2+3k+6\), \(3k^2+4k+4\) are three consecutive terms of an AP.

Split 207 into three parts in AP such that the product of the two smaller parts is 4623.

The angles of a triangle are in AP; the greatest is twice the least. Find all angles.

If the \(n\)th terms of APs \(9,7,5,\ldots\) and \(24,21,18,\ldots\) are equal, find \(n\) and that term.

If \(a_3+a_8=7\) and \(a_7+a_{14}=-3\), find \(a_{10}\).

Find the 12th term from the end of AP: \(-2,-4,-6,\ldots,-100\).

Which term of AP \(53,48,43,\ldots\) is the first negative term?

How many numbers between 10 and 300 leave remainder 3 when divided by 4?

Find the sum of the two middle-most terms of the AP: \(-\dfrac{4}{3}, -1, -\dfrac{2}{3},\ldots, 4\dfrac{1}{3}\).

First term is \(-5\), last term is \(45\), and sum of the AP is \(120\). Find the number of terms and the common difference.

Find the sum:

(i) \(1+(-2)+(-5)+\cdots+(-236)\)

(ii) \(\big(4-\dfrac{1}{n}\big)+\big(4-\dfrac{2}{n}\big)+\cdots+\big(4-\dfrac{n}{n}\big)\)

(iii) \(\dfrac{a-b}{a+b}+\dfrac{3a-2b}{a+b}+\dfrac{5a-3b}{a+b}+\cdots\) to 11 terms

Which term of the AP \(-2,-7,-12,\ldots\) is \(-77\)? Find the sum up to that term.

If \(a_n=3-4n\), show \(a_1,a_2,a_3,\ldots\) form an AP and find \(S_{20}\).

In an AP, if \(S_n=n(4n+1)\), find the AP.

In an AP, if \(S_n=3n^2+5n\) and \(a_k=164\), find \(k\).

If \(S_n\) is the sum of first \(n\) terms of an AP, prove \(S_{12}=3(S_8-S_4)\).

Find \(S_{17}\) if \(a_4=-15\) and \(a_9=-30\).

If \(S_6=36\) and \(S_{16}=256\), find \(S_{10}\).

Find the sum of all 11 terms of an AP whose middle-most term is 30.

Find the sum of the last 10 terms of the AP: \(8,10,12,\ldots,126\).

Find the sum of first seven numbers which are multiples of 2 as well as of 9.

How many terms of the AP: \(-15,-13,-11,\ldots\) are needed to make the sum \(-55\)? Explain the double answer.

The sum of first \(n\) terms of an AP with \(a=8\), \(d=20\) equals the sum of first \(2n\) terms of another AP with \(a=-30\), \(d=8\). Find \(n\).

Kanika deposits Rs 1 on day 1, Rs 2 on day 2, … through January 2008 (31 days). She also spent Rs 204 and still had Rs 100 left. What was her pocket money for the month?

Yasmeen saves Rs 32, Rs 36, Rs 40 in months 1,2,3 and continues so (AP with \(d=4\)). In how many months will she save Rs 2000?