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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