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# The sum of the third and seventh term of an A.P. is 6 and their product is 8. Find the sum of the first sixteen terms of the A.P. Verified
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Hint: In problems of A.P., we consider the first term of the A.P. be $a$ and common difference be $d$. Let the number of A.P. be $n$.
Then we apply the formula
${{T}_{n}}=a+(n-1)d$ to get the $n\text{th}$ term of A.P and ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ to find sum of first $n$ terms of A.P.

In this question Arithmetic Progression also known as A.P. will be used. An A.P. is a sequence of numbers such that the difference of any two successive numbers is a constant called common difference of the A.P. An A.P. is generally represented as:

$a,\text{ }a+d,\text{ }a+2d,\text{ }a+3d,\text{ }.........$, where $a$ is the first term and $d$ is the common difference.

To find the $n\text{th}$ term we use ${{T}_{n}}=a+(n-1)d$ and to find sum of first $n$ terms

we use ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.

Now, we have been given that:

\begin{align} & {{T}_{3}}+{{T}_{7}}=6 \\ & \therefore a+\left( 3-1 \right)d+a+\left( 7-1 \right)d=6 \\ & \therefore 2a+8d=6 \\ & \therefore a+4d=3 \\ & \therefore a=3-4d..........................................\left( 1 \right) \\ \end{align}

Also we are given that:

\begin{align} & {{T}_{3}}\times {{T}_{7}}=8 \\ & \therefore \left( a+2d \right)\times \left( a+6d \right)=8.........................\left( 2 \right) \\ \end{align}

Substituting the value of $a$ from equation $\left( 1 \right)$ in equation $\left( 2 \right)$,

we get

\begin{align} & \left( 3-4d+2d \right)\times \left( 3-4d+6d \right)=8 \\ & \therefore \left( 3-2d \right)\left( 3+2d \right)=8 \\ \end{align}

Using the formula that: $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, we get

\begin{align} & {{3}^{2}}-{{\left( 2d \right)}^{2}}=8 \\ & \therefore 9-4{{d}^{2}}=8 \\ & \therefore 4{{d}^{2}}=1 \\ & \therefore {{d}^{2}}=\dfrac{1}{4} \\ & \therefore d=\pm \sqrt{\dfrac{1}{4}} \\ & \therefore d=\pm \dfrac{1}{2}. \\ \end{align}

From equation $\left( 1 \right)$, when $d=\dfrac{1}{2}$, $a=1$.

Also from equation $\left( 1 \right)$, when $d=\dfrac{-1}{2}$, $a=5$.

Now, we have to find ${{S}_{16}}$.

\begin{align} & \because {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\ & \therefore {{S}_{16}}=\dfrac{16}{2}\left[ 2a+\left( 16-1 \right)d \right] \\ & \therefore {{S}_{16}}=8\times \left[ 2a+15d \right]. \\ \end{align}

When $a=1$ and $d=\dfrac{1}{2}$, we have

\begin{align} & {{S}_{16}}=8\times \left[ 2\times 1+15\times \dfrac{1}{2} \right] \\ & \therefore {{S}_{16}}=8\times \left[ 2+\dfrac{15}{2} \right] \\ & \therefore {{S}_{16}}=8\times \left[ \dfrac{4+15}{2} \right] \\ & \therefore {{S}_{16}}=8\times \dfrac{19}{2} \\ & \therefore {{S}_{16}}=36. \\ \end{align}

When $a=5$ and $d=\dfrac{-1}{2}$, we have

\begin{align} & {{S}_{16}}=8\times \left[ 2\times 5+15\times \dfrac{-1}{2} \right] \\ & \therefore {{S}_{16}}=8\times \left[ 10-\dfrac{15}{2} \right] \\ & \therefore {{S}_{16}}=8\times \left[ \dfrac{20-15}{2} \right] \\ & \therefore {{S}_{16}}=8\times \dfrac{5}{2} \\ & \therefore {{S}_{16}}=20. \\ \end{align}

Hence, ${{S}_{16}}=36$ when $a=1$ and $d=\dfrac{1}{2}$ or ${{S}_{16}}=20$when $a=5$
and $d=\dfrac{-1}{2}$.

Note: Here, we are getting two values of $d$ and corresponding two values of $a$. Don’t get confused as to which value we have to take, because both values are real and we have to do calculation for both the values separately, just as we did here.

Last updated date: 18th Sep 2023
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