Question

the sum of 3 terms in Ap is 21 and product of 1st and 3rd term exceeds the 2nd term by 6 find the 3 terms

Answer 1

Hi mate !!

Here's the answer !!

Let the three terms be a₁, a₂ and a₃.

Given that,

= a₁ + a₂ + a₃ = 21

= a + a + d + a + 2d = 21

= 3a + 3d = 21

=> a + d = 7   -----( Eqn. 1 )

But a + d = a₂ 

=> a₂ = 7

Also, 

= ( a ) ( a + 2d ) = a₂ + 6   -----( Eqn. 2 )

From Eqn 1 we get a = 7 - d

Substitute that in Eqn. 2

= ( 7 - d ) ( 7 - d + 2d ) = 7 + 6

= ( 7 )² - d² = 13

= 49 - d² = 13

=> -d² = 13 - 49

= - d² = -36

=> d² = 36

Taking square root on both sides, we get,

d = 6

=> a = 7 - d

= a = 7 - 6 = 1

Hence a = 1

 a₂ = 7

a₃ = 7 + 6 =13

Hence the three terms are 1, 7 , 13.

Hope my answer helped !!

Cheers !!