The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6,
Question:
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
Solution:
Let the three terms of the A.P. be $a-d, a, a+d$.
Then, we have:
$a-d+a+a+d=21$
$\Rightarrow 3 a=21$
$\Rightarrow a=7 \ldots(i)$
Also, $(a-d)(a+d)-a=6$
$\Rightarrow a^{2}-d^{2}-a=6$
$\Rightarrow 49-d^{2}-7=6$
$\Rightarrow 36=d^{2}$
$\Rightarrow \pm 6=d$
When $\mathrm{d}=6, \mathrm{a}=7$, we get:
$1,7,13$
When $\mathrm{d}=-6$, a $=7$, we get:
$13,7,1$
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