Question

For an $A.P.$ if $T_n=6n+5$ then find $S_n$.

Answer 1

Given: $T_n=6n+5$ .... $\left(1\right)$

$T_n=a+(n-1)d$, where $a=$ First term, $d=$ Difference between terms ... $\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$

$6n+5=a+(n-1)d$

$6n+5=a+nd-d$ ..... $\left(3\right)$

Equating n terms:$6n=nd$

$d=6$ ...... $\left(4\right)$

Substituting $\left(4\right)$ in $\left(3\right)$

$6n+5=a+6n-6$

$5=a-6$

$a=11$ ..... $\left(5\right)$

Now

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

$S_n=\frac{n}{2}(a+T_n)$ ....... $\left(6\right)$

Substituting $\left(1\right)$ and $\left(5\right)$ in $\left(6\right)$

$S_n=\frac{n}{2}(11+6n+5)$

$S_n=\frac{n}{2}(16+6n)$

$S_n=n(8+3n)$

$S_n=3n^2+8n$