Question

if the sum of first m terms of an AP is n and the sum of first n terms is m, then shiw that the sum of first (m+n) terms is -(m+n).

Answer 1

◢LET'S

●FIRST TERM OF AP = a

◢AND

●COMMON DIFFERENCE = d
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◢THEN

$= > s _{m} = n \: \: \: \: \: \: ...[GIVEN] \\ \\ = > \frac{m}{2} [2a + (m - 1)d] = n \\ \\ = > 2n = 2ma + m(m - 1)d \: \: \: \: \: \: ...[Eq _{1}]$
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◢AND

$= > s _{n} = m \: \: \: \: ..[GIVEN] \\ \\ = > \frac{n}{2} [2a + (n - 1)d] = m \\ \\ = > 2m = 2an + n(n - 1)d \: \: \: \: ..[Eq _{2}]$
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●SUBSTRACTING [Eq(2) - Eq(1)]

=> 2m-2n = [2an+n(n-1)d] - [2am+m(m-1)d]

=> 2m-2n = [2an-2am] + [n(n-1)d-m(m-1)d]

=> 2m-2n = 2a[n-m] + d[n²-n-m²+m]

=> 2m-2n = 2a[n-m] + d[(n²-m²)-(n-m)]

=> -2[n-m] = 2a[n-m] + d[(n-m)(n+m) - (n-m)]

=> -2 = 2a + d[n+m-1]

=> 2a + d[n+m-1] = -2. ________[◢Eq(3)]

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◢NOW

$= > s _{m + n} = \frac{m + n}{2} [2a + (m + n - 1)d] \\ \\ = > s _{m + n} = \frac{m + n}{2} ( - 2) \: \: \: \: \: \: \: ..[By \: Eq _{3}] \\ \\ = > s _{m + n} = - (m + n)$
_____________________[◢PROVED]

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