Question

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

Answer 1

Answer:

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 .......

Answer 2

Hey there!

Let the three terms of A.P be (a-d) , a , (a+d).

According to question,

(a - d) + a + (a + d) = 33

3a = 33

a = 11   ............( i)

And,

(a + d) ( a - d) = a + 29

a² - d² = a + 29

(11)² - d² = 11 + 29             [from (i)]

121 - 11 - 29 = d²

81 = d²

or, d = ± 9

Now,

A.P series is given as: [ when d = 9]

a , a + d , a + 2d , a + 3d, .............

11, 11 + 9 , 11 + 2*9, 11 + 3*9,...........

11 , 19 , 29 , 38, .............