Test Index

Sequence and Series

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Question : 27 of 86

Marks: +1, -0

If the sum of first

n

terms of an A.P.'s Zero, then the sum of next

m

a

is the first term of the A.P. is

$\; \frac{-a m}{(m+n)(n-1)}$
$\; \frac{-a m(m+n)}{n-1}$
$\; \frac{a m(m+n)}{n-1}$
None

Solution:

Concept:

Sum of next

m

terms = total sum of first

(n+m)

terms minus sum of first

n

terms.

Explanation:

Let the first term be

a

and common difference

d

.

Sum of first

n

terms:

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

.

Since

n \neq 0

, we get

2a + (n-1)d = 0

, so

d = \frac{-2a}{n-1}

.

Sum of first

(n+m)

terms:

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

.

Substitute

d

:

S_{n+m} = \frac{n+m}{2} \left[ 2a + (n+m-1)\left(\frac{-2a}{n-1}\right) \right]

.

Simplify:

S_{n+m} = a(n+m) \left[ 1 - \frac{n+m-1}{n-1} \right] = a(n+m) \left[ \frac{(n-1)-(n+m-1)}{n-1} \right] = a(n+m) \left( \frac{-m}{n-1} \right) = \frac{-a m (n+m)}{n-1}

.

Since

S_n = 0

, the sum of next

m

terms equals

S_{n+m}

.

Answer:

\frac{-a m(m+n)}{n-1}

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