Question

Find five terms in A.P. whose sum is $12\frac{1}{2}$ and the ratio of first to the last term is $2:3$.

Answer 1

Sum of an AP is give by

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

Since sum of first 5 terms is given so

$S_5=\frac{5}{2}[2a+(5-1\left)d\right]$

$\frac{25}{2}=\frac{5}{2}[2a+4d]$

$a+2d=\frac{5}{2}..\left(1\right)$

Also given,

$\frac{a}{d}=\frac{2}{3}$

$a=\frac{2d}{3}......\left(2\right)$

Putting value of a from (2) to (1) we get

$\frac{2d}{3}+2d=\frac{5}{2}$

$\frac{2d+6d}{3}=\frac{5}{2}$

$4d+12d=15$

$16d=15$

$d=\frac{15}{16}$

Putting value of d in (2)

$a=\frac{2d}{3}=\frac{2}{3}\times \frac{15}{16}=\frac{5}{8}$

Therefore, first five terms of an AP are $a,a+d,a+2d,a+3d,a+4d$

i.e. $\frac{5}{8},\frac{25}{16},\frac{40}{16},\frac{55}{16},\frac{70}{16}$