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Question
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
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Solution
nth term of an A.P. = tn = 8n – 5
Let a be the first term and d be the common difference of this A.P.
Then,
a = t1
= 8 × 1 – 5
= 8 – 5
= 3
t2 = 8 × 2 – 5
= 16 – 5
= 11
∴ d = t2 – t1
= 11 – 3
= 8
The sum of n terms of an A.P. = `S = n/2 [2a + (n - 1)d]`
`=>` Sum of 28 terms of an A.P. = `28/2 [2 xx 3 + 27 xx 8]`
= 14[6 + 216]
= 14 × 222
= 3108
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