How many terms of the $A.P.$ $-6,-\frac{11}{2},-5, \ldots \ldots$ are needed to give the sum $-25 ?$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Let the sum of $n$ terms of the given $A.P.$ be $-25$

It is known that,

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

Where $n=$ number of terms, $a=$ first term, and $d=$ common difference

Here, $a=-6$

$d=-\frac{11}{2}+6=\frac{-11+12}{2}=\frac{1}{2}$

Therefore, we obtain

$-25=\frac{n}{2}\left[2 \times(-6)+(n-1)\left(\frac{1}{2}\right)\right]$

$\Rightarrow-50=n\left[-12+\frac{n}{2}-\frac{1}{2}\right]$

$\Rightarrow-50=n\left[-\frac{25}{2}+\frac{n}{2}\right]$

$\Rightarrow-100=n(-25+n)$

$\Rightarrow n^{2}-25 n+100=0$

$\Rightarrow n^{2}-5 n-20 n+100=0$

$\Rightarrow n(n-5)-20(n-5)=0$

$\Rightarrow n=20$ or $5$

Similar Questions

The mean of the series $a,a + nd,\,\,a + 2nd$ is

If ${A_1},\,{A_2}$ be two arithmetic means between $\frac{1}{3}$ and $\frac{1}{{24}}$ , then their values are

If the ${n^{th}}$ term of an $A.P.$ be $(2n - 1)$, then the sum of its first $n$ terms will be

Let the sequence $a_{n}$ be defined as follows:

${a_1} = 1,{a_n} = {a_{n - 1}} + 2$ for $n\, \ge \,2$

Find first five terms and write corresponding series.

Four numbers are in arithmetic progression. The sum of first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is