Question

The sum of the 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 3 terms be (a-d), a and (a+d)
a-d+a+a+d=33
3a=33
a=11
(a-d)(a+d)=a+29
a²-d²=a+29
121-d²=11+29
121-d²=40
d²=81
d=+/-9

When d=+9
The terms are 2, 11 and 20

When d=-9
The terms are 20,11 and 2

Answer 2

Let the first term be a ,

The common difference be d .

Then the sum of first three terms = a+a+d+a+2d = 3a+3d .

Given 3(a+d) = 33

=> a+d = 11 .

=> d =11-a

Therefore ,Second term of A.P = 11 .

The product of first and third terms = (a)(a+2d) = a(a+2(11-a)

= a(a+22-2a)

= a(22-a)

= 22a-a²

ATQ --->

Given 22a-a²-29= a+d

=> 22a-a²-29=11

=> 22a-a² -40 =0

=> a²-22a+40=0

=> a²-20a-2a+40=0

=> a(a-20)-2(a-20) =0

=> a= 2 or 20.

Finding common difference for a = 2

11-2=9

Finding Common difference for a =20

11-20=-9 .

Now The possible A .P 's are

1) 2,11,20,29,38,47,56,65,74.....

2) 20,11,2,-7,-16,-25,-34,-43,-52,-61,-70 .......