The fifth term of an A.P. is 1 whereas its 31st term is -77. Find its 20th term and sum of its first fifteen terms. Also find which term of the series will be - 17 and sum of how many terms will be 20.

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Solution

Given $T_{5} = a + 4 d = 1, T_{31} = a + 30 d = - 77,$
Solving the above
two, we get $a = 13$ and $d = - 3.$
$T_{20} = 13 - (20 - 1) 3 = - 44$
$S_{15} = [n / 2] [2 a + [n - 1] d] = [15 / 2] [26 + 14 [- 3]] = - 120$
Let $T_{n} = - 17.$
Then $a + [n - 1] d = - 17.$ $13 + [n - 1] [- 3] = 17 ∴ 3 n = 33$
or $n = 11$
Let $S_{n} = 20$ Then $(n / 2) [2 a + (n - 1) d] = 20.$
$n [n \times 13 + (n - 1) (- 3) = 40]$ by $[1] .$
$3 n^{2} - 29 n + 40 = 0 ∴ [n - 8] [3 n - 5] = 0 ∴ n = 8$
The value of n cannot be fractional.

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