Question

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> The sum of the third and the seventh terms of an $AP$ is $6$ and their
product is $8$. Find the sum of first sixteen terms of the $AP$ .

Answer 1

From the given statements, we can write,
$a_3+a_7=6...............\left(1\right)\text{and}$
$a_3+a_7=8...............\left(2\right)$
By the $n^{th}$ formula,
$a_n=a+(n+1)d$
Third term, $a_3=a+(3-1)d=a+2d............\left(3\right)\text{and}$
Seventh term $a_7=a+(7-1)d=a+6d............\left(4\right)$

Putting equation (3) an (4) in equation (1) we get,
$a+2d+a+6d=6$
$2a+8d=6$
$a+4d=3$
$a=3-4d............................\left(5\right)$

Again putting equation (3) an (4) in equation (2), we get
$a+2d\left)x\right(a+6d)=8$
Substituting the value of a from equation $\left(5\right)$, we get,
$3-4d+2d\left)x\right(3-4d+6d)=8$
$(3-2d)x(3+2d)=8$
$3^2-2d^2=8$
$9-4d^2=8$
$9-8=4d^2$
$1=4d^2$
$d=\frac{1}{2}!\text{or}-\frac{1}{2}$
Now, by putting both the value of $d,$ we get,
$a=3-4d=3-4\left(\frac{1}{2}\right)=3-2=1,\text{when}d=\frac{1}{2}$
$a=3-4d=3-4(-\frac{1}{2})=3+2=5,\text{when}d=-\frac{1}{2}$

We know the sum of $n^{th}$ term of $A.P$ is
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

so when $a=1\text{and}d=\frac{1}{2}$
Then, the first $16$ terms are

${\S }_{16}=\frac{16}{2}[2+(16-1\left)\frac{1}{2}\right]=8(2+\frac{15}{2})=76$

And when the $a=5$ and $d=-\frac{1}{2}$
$S_{16}=\frac{16}{2}[2.5+(16-1)\left(\frac{1}{2}\right)]=8\left(\frac{5}{2}\right)=20$