Question

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

Step 1: Find the sum of the first sixteen terms of the AP

From the given data,

$a_3+a_7=6\dots \dots \dots .\left(i\right)$

Also,

$a_3\times a_7=8\dots \dots \dots ..\left(ii\right)$

We know the $nth$ term formula,

$a_n=a+(n-1)d$

Third term, $a_3=a+(3-1)d$

$a_3=a+2d\dots \dots \dots \dots \dots \dots \dots \left(iii\right)$

And Seventh term, $a_7=a+(7-1)d$

$a_7=a+6d\dots \dots \dots \dots \dots ..\left(iv\right)$

From equation (iii) and (iv), putting in equation(i), we get,

$\begin{array}{rcl}a+2d+a+6d& =& 6\\ 2a+8d& =& 6\\ a+4d& =& 3\end{array}$

or

$a=3–4d\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(v\right)$

Step 2: Again putting the eq.(iii) and (iv), in eq. (ii)

we get,

$(a+2d)\times (a+6d)=8$

Putting the value of a from equation (v), we get,

$\begin{array}{rcl}(3–4d+2d)\times (3–4d+6d)& =& 8\\ (3–2d)\times (3+2d)& =& 8\\ 3^2–2d^2& =& 8\\ 9–4d^2& =& 8\\ 4d^2& =& 1\\ d& =& \frac{1}{2}or-\frac{1}{2}\end{array}$

Step 3: Now, by putting both the values of $d$

we get,

$$\begin{gathered} a=3–4d \\ a=3–4\left(\frac{1}{2}\right) \\ a=3–2 \\ a=1 \end{gathered}$$, when $d=\frac{1}{2}$

$$\begin{gathered} a=3–4d \\ a=3–4(-\frac{1}{2}) \\ a=3+2 \\ a=5 \end{gathered}$$, when $d=-\frac{1}{2}$

We know, the sum of nth term of AP is;

$Sn=\frac{n}{2}[2a+(n–1\left)d\right]$

So, when $a=1$ and $d=\frac{1}{2}$

Step 4: Now let us find the sum of first sixteen terms

Then, the sum of first $16$ terms are;

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

And when $a=5$ and $d=-\frac{1}{2}$

Step 5: Then, find the sum of first $16$ terms

$$\begin{gathered} S_{16}=\frac{16}{2}[2.5+(16-1\left)\right(-\frac{1}{2}\left)\right] \\ {S}_{16}=8\left(\frac{5}{2}\right) \\ {S}_{16}=20 \end{gathered}$$

Hence, the sum of first sixteen terms of the AP is $76$ or $20$.