Find the sum of first 25 terms of an AP whose nth term is 1 - 4n.

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Solution

Given,

$n^{t h}$ term $T_{n} = 1 - 4 n$

put, $n = 1$ , $T_{1} = a = 1 - 4 (1) = - 3$

put, $n = 2$ , $T_{2} = 1 - 4 (2) = - 7$

Common difference, $d = T_{2} - T_{1} = - 7 - (- 3) = - 4$

So, we have $a = - 3, d = - 4, n = 25$ , to find the sum.

Sum, $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$

$S_{25} = \frac{25}{2} (2 (- 3) + (25 - 1) (- 4))$

$S_{25} = \frac{25}{2} (- 112) = 25 \times (- 51)$

$S_{25} = - 1275$

This is the sum.

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