Question

if sum of first m terms of an AP is the same as the sum of its first n terms.show that the sum of it's first (m+n) terms is zero.

Answer 1

according to question

m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)

cutting 2

we get

m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

2am + m^2d - md -2an -n^2d +nd =0

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

taking (m-n) common

2a + ( m + n -1) d = 0------------ 1

S m+n = m+n/2( 2a + (m+n -1)d

we know that 2a + (m+n)d is 0 from eqn. 1

therefore S m+n = 0

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Answer 2

Solution:

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Given:

Sum of first m terms of an AP is the same as the sum of its first n terms,.

$S_m = S_n$

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To show :

Sum of its first (m+n) terms is 0.

=> $S_{m+n} = 0$

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Proof:

As we know that,

$S_n = \frac{n}{2} (2a + (n-1)d)$

So,

=> $\frac{n}{2} (2a + (n-1)d) = \frac{m}{2} (2a +(m-1) d)$

=> $n(2a +nd -d) = m (2a + md -d)$

=> $2an + n^2d -nd = 2am +m^2d -md$

=> $2an - 2am + n^2d -m^2d -nd + md = 0$

=> $2a (n-m) + (n^2-m^2)d -(n -m)d = 0$

=> $2a(n-m) + (n-m)(n+m)d - (n-m)d = 0$

=> $(n-m)(2a +(n+m)d - d) = 0$

=> $2a + (n+m-1)d =0$ ...(i)

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As, we know

$S_n = \frac{n}{2} (2a + (n-1)d)$

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

=> $S_{m+n} = \frac{m+n}{2} (0)$

=> $S_{m+n} = 0$

∴ Hence proved

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