The sum of three terms which are in A.P is $33$, if the product of the $1st$ and $3rd$ terms exceeds the $2nd$ term by $29$, find AP.

Given: The sum of three terms which are in A.P is $33$, if the product of the $1st$ and $3rd$ terms exceeds the $2nd$ term by $29$.

To do: To find AP.  

Solution:

Let the first three terms of AP is $a−d,\ a,\ a+d$.

According to problem,

$a−d+a+a+d=33\ ......( 1)$

$( a−d)( a+d)=a+29\ .....( 2)$

From $( 1)$, we get

$3a=33$

$\Rightarrow a=\frac{33}{3}=11$

From $( 2)$, we get

$a^2−d^2=a+29\  .....( 3)$

On putting $a=11$ in equation $( 3)$

$121−d^2=11+29$

$\Rightarrow 121−d^2=40$

$\Rightarrow 121−40=d^2$
 
$\Rightarrow 81=d^2$
 
$\Rightarrow d=±9$

When $a=11$ & $d=9$

A.P: $2,\ 11,\ 20,\ 29,\ .....$

When $a=11$ & $d=−9$

A.P: $20,\ 11,\ 2,\ ......$
 

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Updated on: 10-Oct-2022

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