Question

The fourth term of an Arithmetic progression is $10$ and the eleventh term of it exceeds three times the fourth term by $1.$ Find the sum of the first $20$ terms of the progression.

Answer 1

Given fourth term of A.P is $10$.
$\Rightarrow a_4=10\to \left(1\right)$, where, $a_4$ is denoted by fourth term of an A.P .
And also eleventh term exceeds three times the fourth term by $1$.
$\Rightarrow a_{11}=3a_4+1\to \left(2\right)$
then we should find $S_{20}=?$

We know, $a_n=a_1+(n-1)d$

Using this formula in $\left(1\right)$ and $\left(2\right)$,

$a+3d=10\to \left(3\right)$

$a+10d=3(a+3d)+1$
$\Rightarrow a+10d=3a+9d+1$
$\Rightarrow 10d-9d=3a-a+1$

$⟹d=2a+1\to \left(4\right)$

Putting $\left(4\right)$ in $\left(3\right)$,

$\Rightarrow a+3(2a+1)=10$
$\Rightarrow a+6a+3=10$
$\Rightarrow 7a+3=10$
$\Rightarrow 7a=10-3$
$\Rightarrow a=\frac{7}{7}$

$\therefore a=1$
Substitute $a=1$ in equation(4), we get

Thus, $d=3\left(1\right)+1$
$=3+1$
$\therefore d=4$

Now, $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$
Substitute $n=20,a=1,d=4$ in above formula, we get

$S_{20}=\frac{20}{2}(2\times (1)+(20-1)\times 3)$
$=\frac{20}{2}(2+(19\times 3\left)\right)$
$=10(2+57)$
$=10\left(59\right)$

$\therefore S_{20}=590$