Question

★Question How many terms of the AP -1,-5,-9 must be taken to get a sum of -496​

Answer 1

Given :-

The AP of - 1, - 5, - 9 must be taken to get a sum of - 496.

To Find :-

How many numbers of terms can be found in an AP.

Formula Used :-

$\clubsuit$ Sum of first n terms of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{S_n =\: \dfrac{n}{2}\bigg\lgroup 2a + (n - 1)d\bigg\rgroup}}}$

where,

$\sf S_n$ = Sum of first n terms of an AP

a = First term of an AP

a = First term of an APd = Common difference

a = First term of an APd = Common differencen = Number of terms of an AP

Solution :-

Solution :-Given :

$\bigstar\: \: \: \bf{First\: term\: of\: an\: AP\: (a) =\: - 1}$

$\bigstar\: \: \: \bf{Common\: difference\: (d) =\: - 4}$

$\bigstar\: \: \: \bf{Sum\: of\: n\: terms\: (S_n) =\: - 496}$

According to the question by using the formula we get,

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup 2(- 1) + (n - 1)(- 4)\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup -2 + (n - 1)(- 4)\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup - 2 - 4(n - 1)\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup - 2 - 4n + 4\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup - 2 + 4 - 4n\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup 2- 4n\bigg\rgroup$

$\implies \sf - 496 =\: n(1 - 2n)$

$\implies \sf - 496 =\: n - 2n^2$

$\implies \sf - 496 - n + 2n^2 =\: 0$

$\implies \sf 2n^2 - n - 496 =\: 0$

$\implies \sf 2n^2 - (32 - 31)n - 496 =\: 0$

$\implies \sf 2n^2 - 32n + 31n - 496 =\: 0\: \: \bigg\lgroup \small\sf\bold{\pink{By\: splitting\: middle\: term}}\bigg\rgroup$

$\implies \sf 2n(n - 16) + 31(n - 16) =\: 0$

$\implies \sf (2n + 31)(n - 16) =\: 0$

$\implies \bf{(2n + 31) =\: 0}$

$\longrightarrow \sf 2n + 31 =\: 0$

$\longrightarrow \sf 2n =\: - 31$

$\longrightarrow \sf\bold{\purple{n =\: \dfrac{- 31}{2}}}\: \: \bigg\lgroup \small\sf\bold{\pink{Number\: of\: terms\: can't\: be\: negetive (- ve)}}\bigg\rgroup$

$\implies \bf{(n - 16) =\: 0}$

$\longrightarrow \sf n - 16 =\: 0$

$\longrightarrow \sf \bold{\red{n =\: 16}}$

$\therefore$ The number of terms can be found in an AP is 16.