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Question
If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.
Solution
T3 = 1, T6 = –11, n = 32
a + 2d = 1 ...(i)
a + 5d = –11 ...(ii)
– – +
Subtracting (i) and (ii),
–3d = 12
⇒ d = `(12)/(-3)` = –4
Substitute the value of d in eq. (i)
a + 2(–4) = 1
⇒ a – 8 = 1
a = 1 + 8 = 9
∴ a = 9, d = –4
S32 = `n/(2)[2a + (n - 1)d]`
= `(32)/(2)[2 xx 9 + (32 - 1) xx (–4)]`
= 16[18 + 31 x (–4)]
= 16[18 – 124 ]
= 16 x (–106)
= –1696.
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