The sum of three terms of an A.P. is 21 and the product

In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.

We need to find the three terms.

Here,

Let the three terms be $(a-d), a,(a+d)$ where, $a$ is the first term and $d$ is the common difference of the A.P

So.

$(a-d)+a+(a+d)=21$

$3 a=21$

$a=7$  ................(1)

Also,

$(a-d)(a+d)=a+6$

$a^{2}-d^{2}=a+6$ (Using $a^{2}-b^{2}=(a+b)(a-b)$ )

$(7)^{2}-d^{2}=7+6$ (Using 1)

$49-13=d^{2}$

Further solving for d,

$d^{2}=36$

$d=\sqrt{36}$

$d=6 \quad \ldots \ldots(2)$

Now, using the values of a and d in the expressions of the three terms, we get,

First term $=a-d$

So,

$a-d=7-6$

$=1$

Second term = a

So,

$a=7$

Also,

Third term $=a+d$

So,

$a+d=7+6$

$=13$

Therefore, the three terms are 1,7 and 13 .