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Question
If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero
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Solution
Let a be the first term and d be the common difference of the AP.
It is given that the sum of first m terms is same as the sum of its first n terms.
∴Sm = Sn
⇒`m/2`[2a +(m − 1)d] = `n/2`[2a + (n − 1)d]
⇒2am +m(m − 1)d = 2an + n(n − 1)d
⇒2a(m − n) =[(n2 − n)−(m2 - m)]d
⇒2a(m − n) =[(m − n) − (n − m)(n + m)]d
⇒2a(m − n) = −(m − n)(−1 + m +n)d
⇒2a = −(m + n − 1)d .....(1)
Now,
Sum of first (m + n) terms
`= S_"m+n"`
`= (m+2)/2 [2a + (m + n - 1)d]`
`=(m+2)/2 [-(m+n-1)d + (m+n-1)d]` [From 1]
`= "m+n"/2 xx 0`
= 0
Thus, the sum of first (m + n) terms of the AP is zero.
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