Question

Find the first term in A.P., whose sum is $50$ and in which the greatest number is $4$ times the least.

Answer 1

Let the four numbers are $T_1,T_2,T_3,T_4$

Given that They are in $A.P$

then $T_1=a$

$T_2=a+d$

$T_3=a+2d$

$T_4=a+3d$

Where $d$ is the common difference

Let First term is the smallest number

Now Given $T_1+T_2+T_3+T_4=50$

$\Rightarrow a+(a+d)+(a+2d)+(a+3d)=50$

$\Rightarrow 4a+6d=50$

$2a+3d=25.....\left(1\right)$

Ans Also given

$T_4=4\left(T_1\right)$

$a+3d=4a$

$d=a......\left(2\right)$

From eqn $\left(1\right)$ and $\left(2\right)$

$2a+3a=25$

$a=5$

Hence $T_1=a=5$

$T_2=a+d=10$

$T_3=a+2d=15$

$T_4=a+3d=20$