Question

The sum of first m terms of ap is equal to the sum of first n terms of ap. Prove that sum of first (m+n) terms will be 0.

Answer 1

since sum of m terms of an Ap is equal to the sum of n terms of an AP,

therefore, Sm = Sn

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

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

2am + dm^2 - dm = 2an + dn^2 - dn

2am - 2an + dm^2 - dn^2 - dm + dn = 0

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

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

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

2a + d(m+n-1) = 0  - (1)

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

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

by equation (1)

(m+n)/2 [0] = 0

therefore, S(m+n) = 0

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