Taking the first term of an A.P. to be α, the sum of first m terms is zero, then the sum of next n terms will be

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RPSC 2nd Grade Mathematics (Held on 18th Feb 2019) Official Paper
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  1. \(\frac{n\alpha(n−m)}{m−1}\)
  2. \(\frac{n\alpha(n+m)}{m−1}\)
  3. \(\frac{n\alpha(n−m)}{n−1}\)
  4. \(\frac{n\alpha(m+n)}{1−m}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{n\alpha(m+n)}{1−m}\)
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Detailed Solution

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Concept used:

Sum of first n terms of A.P

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

Calculation:

First term = a

S= 0 

∴ m/2[2a + (m - 1)d] = 0

d = -2a/(m - 1) ---(i)

am + 1 = a + md

And md will be the first term for next n terms

S n/2[2a + (n - 1)d]

Sn =  n/2[2(a + md) + (n - 1)d]

⇒ n/2[2a + 2md + (n - 1)d]

⇒ n/2[2a + (2m + n - 1)d]

Put d from (i),

⇒ n/2[2a + (2m + n - 1)(-2a/(m - 1))]

n/2 × 2a [1 - (2m + n - 1)/(m - 1)]

⇒ an[(m - 1 - 2m - n + 1)/(m - 1)]

∴ Sm = an(n + m)/(1 - m)

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