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Question
If the term of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its (m + n) terms is zero
Sum
Solution
Let a be the first term and d be the common difference of the given A.P. Then,
Sm = Sn
`⇒ \frac { m }{ 2 } [2a + (m – 1) d] = \frac { n }{ 2 } [2a + (n – 1) d]`
⇒ 2a(m – n) + {m (m – 1) – n (n – 1)} d = 0
⇒ 2a (m – n) + {(m2 – n2) – (m – n)} d = 0
⇒ (m – n) [2a + (m + n – 1) d] = 0
⇒ 2a + (m + n – 1) d = 0
⇒ 2a + (m + n – 1) d = 0 [∵ m – n ≠ 0] ….(i)
`S_(m+n)=(m+n)/2{2a+(m+n-1)d}`
`S_(m+n)=(m+n)/2 xx 0 = 0`
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