Question

Find the sum of first $20$ terms of the arithmetic series in which $3^{rd}$ term is $7$ and $7^{th}$ term is $2$ more than three times its $3^{rd}$ term

Answer 1

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$

It is given that the $3$rd term of the arithmetic series is $7$ that is $T_3=7$ and therefore,

$T_3=a+(3-1)d\Rightarrow 7=a+2d....\left(1\right)$

Also it is given that the $7$th term is $2$ more than three times its $3$rd term that is

$T_7=(3\times T_3)+2=(3\times 7)+2=21+2=23$

Thus,

$T_7=a+(7-1)d\Rightarrow 23=a+6d....\left(2\right)$

Subtract equation 1 from equation 2:

$(a-a)+(6d-2d)=23-7\Rightarrow 4d=16\Rightarrow d=\frac{16}{4}\Rightarrow d=4$

Substitute the value of $d$ in equation 1:

$a+(2\times 4)=7\Rightarrow a+8=7\Rightarrow a=7-8=-1$

We also know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now to find the sum of first $20$ terms, substitute $n=20,a=-1$ and $d=4$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_{20}=\frac{20}{2}\left[\right(2\times -1)+(20-1\left)4\right]=10[-2+(19\times 4\left)\right]=10(-2+76)=10\times 74=740$

Hence, the sum offirst $20$ termsis $740$.