Question

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

Answer 1

SOLUTION:-

Given:

The sum of first term 25 terms of an A.P. whose nth term is 1-4n.

Explanation:

We have,

${}^{a} n = 1 - 4n$

So,

The putting, n=1

${}^{a} 1 = 1 - 4(1) \\ \\ {}^{a} 1 = 1 - 4 \\ \\ {}^{a} 1 = - 3$

&

Putting, n=2.

${}^{a} 2 = 1 - 4(2) \\ \\ {}^{a} 2 = 1 - 8 \\ \\ {}^{a} 2 = - 7$

•Common difference= a2 -a1

=) -7 -(-3)

=) -7 + 3

=) d= -4.

•First term, a= -3.

Therefore,

Formula of the sum of the A.P;

${}^{S} n = \frac{n}{2}[2a + (n - 1)d]$

So,

${}^{s} 25 = \frac{25}{2} [2( - 3) + (25 - 1)( - 4)] \\ \\ {}^{S}25 = \frac{25}{2} [- 6 + 24 \times ( - 4)] \\ \\ {}^{S} 25 = \frac{25}{2} [- 6 + ( - 96)]\\ \\ {}^{S} 25 = \frac{25}{2} [- 6 - 96]\\ \\ {}^{S} 25 = \frac{25}{2} \times (- 102) \\ \\ {}^{S} 25 = 25 \times ( - 51) \\ \\ {}^{S} 25 = - 1275$

Thus,

The sum of the first 25 terms of an A.P. whose nth term is 1-4n is -1275.

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