Question

The sum of three consecutive terms of an arithmetic progression is $18$ while their product is $120$. Find the three terms in A.P.

Answer 1

Let the $3$ consecutive terms in the A.P. be $a-d,a,a+d$

Thus $a-d+a+a+d=18$ and also $(a-d)\left(a\right)(a+d)=120$ Thus, solving these two equations we get $3a=18$ or $a=6$

Thus putting the value of $a=6$ in the second equation, we get

$(6^2-d^2)6=120$ or $36-d^2=20$

Or $d=+/-4$

Thus the three consecutive terms in the A.P. are $2,6,10$