Question

The ${13}^{th}$ term of an AP is four times its $3^{rd}$ term.If its fifth term is 16,then find the sum of its first ten terms.

Answer 1

In the given AP.,let first term=a and common difference =d.

Then, $T_n=a+(n-1)d$

$\Rightarrow T_{13}=a+(13-1)d=a+12d$

and $T_3=a+(3-1)d=a+2d$

Now, $T_{13}=4T_3$ (Given)

a+12d=4(a+2d)

a+12d=4a+8d

3a=4d

a=$\frac{4}{3}d...\left(i\right)$

Also, $T_5=a+(5-1)d$

$\Rightarrow a+4d=16...\left(ii\right)$

Putting the value of a from eq.(i) in (ii), we get

$\frac{4}{3}d+4d=16$

4d+12d=48

16d=48

d=3

Substituting d=3 in eq.(ii),we get

a+4(3)=16

a=16-12

a=4
$\therefore$ Sum of first ten terms is

$S_{10}=\frac{n}{2}[2a+(n-1\left)d\right]$ where n=10

$=\frac{10}{2}[2\times 4+(10-1\left)3\right]$

=5[8+27]

=175