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Question
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
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Solution
Given, t2 = 14 and t3 = 18
`=>` d = t3 – t2
= 18 – 14
= 4
Now t2 = 14
`=>` a + d = 14
`=>` a = 14 – 4
`=>` a = 10
Sum of n terms of an A.P. = `n/2[2a + (n - 1)d]`
∴ Sum of first 51 terms of an A.P. = `51/2 [2 xx 10 + 50 xx 4]`
= `51/2 [20 + 50 xx 4]`
= `51/2 [20 + 200]`
= `51/2 xx 220`
= 51 × 110
= 5610
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