find the required term . .

The given series $39, 78, 117, 156, ...........$ is in A.P . Let the number of terms be $n$

Common difference $d$ = 78 - 39 = 117 - 78 = 156 - 117 = $39$ and the First term $a$ = 39

Now, the sum of $n$ terms in A.P is given by the formula $\frac{n}{2} \times (2a + (n - 1)d)$

Sum of $n$ terms = 4680 = $\frac{n}{2} \times (2 \times 39 + (n-1) \times 39)$

$\Rightarrow 4680 = \frac{n}{2} \times 39(2 + n - 1)$

$\Rightarrow n(n + 1) = \frac{4680 \times 2}{39} = 240$

$\Rightarrow n^2 + n - 240 = 0$

$\Rightarrow (n -15)(n - 16) = 0$

$\Rightarrow n = 15$ (or) $n = -16$ . Since number of terms cannot be negative the number of terms is 15 . And $x$ is the $15^{th}$ term of the given A.P

Now, in an A.P the $n^{th}$ term is given by the formula $T_{n} = a + (n - 1)d$

So, $\color{#20A900}T_{15} = 39 + (15 - 1)39 = 39 + 549 = 585$

Therefore , $\color{#3D99F6}\boxed{x = 585}$

Divide it by 39,and get $1+2+3+4+...+\frac{x}{39}=120$
Let $\frac{x}{39}=y,\frac{y(y+1)}{2}=120$ ,so $y=15,x=585$