Question

The sum of first three terms of an AP is '33. If the product of the first and the third exceeds the second term by 29, find the AP.

Answer 1

Let the first three terms of AP is $a-d,a,a+d$.

According to problem,

$a-d+a+a+d=33$ $\to \left(1\right)$

$(a-d)(a+d)=a+29\to \left(2\right)$

From $\left(1\right)$, we get

$3a=33$

$a=11$

From $\left(2\right)$, we get

$a^2-d^2=a+29\to \left(3\right)$

Put $a=11$ in equation $\left(3\right)$

$121-d^2=11+29$

$121-d^2=40$

$121-40=d^2$

$81=d^2$

$d=\pm 9$

When $a=11$ & $d=9$

A.P: $2,11,20,29,.....$

When $a=11$ & $d=-9$

A.P: $20,11,2,.....$.