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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