Question

The sum of first $n$ terms of an A.P., is zero, show that the sum of next $m$ terms is $\frac{- a m (m + n)}{n - 1}$ , $a$ being the first term.

Accepted Answer

Given:

Sum of first $n$ terms is $0$

Let AP be $a_{1}, a_{2}, a_{3}, . . . . . a_{n}, a_{n + 1}, . . . a_{m + n}$

Thus $a_{1} + a_{2} + a_{3} + . . . . . + a_{n} = 0$

We know that sum of first $n$ terms is

$\Rightarrow S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = 0 \Rightarrow [2 a + (n - 1) d] = 0 \Rightarrow (n - 1) d = - 2 a \Rightarrow d = \frac{- 2 a}{n - 1}$

For next $m$ terms

First term is $a_{n + 1}$

$\Rightarrow a_{n + 1} = a + (n + 1 - 1) d \Rightarrow a_{n + 1} = a + n d$

Let $A$ be the first term of next $m$ term

$∴ A = a + n d$

Now the sum of next $m$ terms

$\Rightarrow S_{m} = \frac{m}{2} [2 A + (m - 1) d] \Rightarrow S_{m} = \frac{m}{2} [2 (a + n d) + (m - 1) d] \Rightarrow S_{m} = \frac{m}{2} [2 (a + n (\frac{- 2 a}{n - 1})) + (m - 1) (\frac{- 2 a}{n - 1})] \Rightarrow S_{m} = \frac{m}{2} \times 2 [(a + n (\frac{- 2 a}{n - 1})) + (m - 1) (\frac{- a}{n - 1})] \Rightarrow S_{m} = m [(a + \frac{- 2 a n}{n - 1}) - \frac{a m}{n - 1} + \frac{a}{n - 1}] \Rightarrow S_{m} = m [(\frac{a (n - 1) - 2 a n}{n - 1}) - \frac{a m}{n - 1} + \frac{a}{n - 1}] \Rightarrow S_{m} = m [(\frac{a n - a - 2 a n}{n - 1}) - \frac{a m}{n - 1} + \frac{a}{n - 1}] \Rightarrow S_{m} = m [\frac{a n - a - 2 a n - a m + a}{n - 1}] \Rightarrow S_{m} = m [\frac{- a n - a m}{n - 1}] \Rightarrow S_{m} = - a m [\frac{m + n}{n - 1}] \Rightarrow S_{m} = \frac{- a m (m + n)}{n - 1}$

Hence proved.

H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Arithmetic Progressions, Exercise 3: EXERCISE

H K Dass Mathematics Solutions for Exercise - H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Arithmetic Progressions, Exercise 3: EXERCISE

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