Question

the sum of three terms which are in an ap is 33 if the product of first term and third term exceeds the second term by 29 find the AP

Answer 1

Let the three terms be a-d, a, a+d
Then a-d+a+a+d=33 or a= 11
Also (a-d)(a+d)=a+29, putting the value of a ,
we get 121-d^2=40 or d^2=81 or d=9 or d=-9
Thus the AP is 2,11,20 or 20,11,2

Answer 2

a+a+d+a+2d=33
3a+3d=33
a+d=11
hence, 2nd term=11
Given that,
a(a+2d)==11+29
a^2+2ad==40
a^2+2ad+d^2=40+d^2
(a+d)^2=40+d^2
11^2=40+d^2
121==40+d^2
d^2=81
d=9
by substituting 'd'Ivoire value in equation1,
a+9=11
a=2
As a=2 & d=9 A.P will be ,
2,11,20,29,..................