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Question
The sum of 5 and 15 terms of an A.P. are equal. Find the sum of 20 terms of this A.P.
Sum
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Solution
We are given that the sum of 5 terms (S5) is equal to the sum of 15 terms (S15)
(S5) = (S15)
`5/2[2a + (4)d] = 15/2[2a + 14(d)]`
2a + 4d = 6a + 42d
−4a = 38d
a = `(-38d)/4`
a = `(-19d)/2`
`S_n = n/2[2a + (n - 1)d]`
`S_20 = 20/2[2((-19d)/2) + (20 - 1)d]`
`S_20 = 10[2((-19d)/2) + 19d]`
S20 = 10[-19d + 19d]
S20 = 10 × 0
S20 = 0
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