Question

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

Answer 1

let the three terms in AP be a-d,a,a+d.
now, by condition,

(a-d)+a+(a+d)=33
a+a+a=33
3a=33
a=11
now, it is also given that product of first and third term exceeds the second term by 29.
now,
(11-d)(11+d)=a+29
121-d^2=11+29
121-d^2=40
d^2=121-40
d^2=81
d=9 or -9.
therefore, the three terms of AP are either 2,11,210or 20,11,2.