Question

Find the value of $n$, if $1+4+7+10+....$ to $n$ terms $=590$.

Answer 1

Given:-

$1+4+7+10+......+n=590$

The above series is in A.P. with first term $1$ and common difference $4-1=3$.

As we know that sum of $n$ terms of an A.P. is given by-

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

Therefore, for the given series,

$S_n=590$

$\frac{n}{2}[2\times 1+(n-1)3]=590$

$\Rightarrow n[2+3n-3]=1180$

$\Rightarrow 3n^2-n=1180$

$\Rightarrow 3n^2-n-1180=0$

$\Rightarrow 3n^2-60n+59n-1180=0$

$\Rightarrow (3n+59)(n-20)=0$

$\Rightarrow n=\frac{-59}{3},20$

Since number of terms cannot be negative.

$\therefore n\ne \frac{-59}{3}$

$\Rightarrow n=20$

Hence the value of $n$ is $20$.