Question

Find five consecutive terms in an $A.P$ such that their sum is $60$ and the product of the third and the fourth term exceeds the fifth by $172$.

Answer 1

Let the terms of AP be

$A-2d,a-d,a,a+d,a+2d$

A/Q

$(a-2d)+(a-d)+a+a(a+d)+(a+2d)=605a=60\Rightarrow a=12$

and $a\times (a+d)=172+a+2d\Rightarrow a^2+ad=172+a+2d\Rightarrow 144+12d=172+12+2d\Rightarrow 12d-2d=184-144=40\Rightarrow 10d=40\Rightarrow d=4\therefore a=12,d=4$

and the terms are

$12-4\times 2,12-4,12,12+4,12+2\times 4=4,8,12,16,20$