# The third term of an A.P. is $18$, and the seventh term is $30$; find the sum of $17$ terms.

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Hint: To find the sum of $17$ terms of the A.P., we have to find the first term and the common

difference of the AP. Using the formula of general term of an AP, we can get two equations by using

the information which is given in the question. Solve these equations to find the first term and the

common terms of the A.P.

Before proceeding in the question, we will first write down all the formulas that are

required to solve this problem.

Let us consider an arithmetic progression having its first term equal to $'a'$ and the common

difference equal to $'d'$ .

The ${{n}^{th}}$ term of this arithmetic progression (denoted by ${{a}_{n}}$) is given by the formula,

${{a}_{n}}=a+\left( n-1 \right)d..........(1)$

The sum of $n$ terms of this arithmetic progression (denoted by ${{S}_{n}}$) is given by the formula,

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

In the question, it is given that the third term i.e. ${{a}_{3}}$ of this arithmetic progression is equal to

18.

$\Rightarrow {{a}_{3}}=18$

If we substitute $n=3$ in equation $\left( 1 \right)$, we get,

${{a}_{3}}=a+\left( 3-1 \right)d$

$\Rightarrow {{a}_{3}}=a+2d$

Substituting this ${{a}_{3}}$ in the equation ${{a}_{3}}=18$, we get,

$a+2d=18..........(3)$

Also, it is given in the question that the seventh term of this arithmetic progression is equal to 30.

$\Rightarrow {{a}_{7}}=30$

If we substitute $n=7$ in equation $\left( 1 \right)$, we get,

${{a}_{7}}=a+\left( 7-1 \right)d$

$\Rightarrow {{a}_{7}}=a+6d$

Substituting this ${{a}_{7}}$ in the equation ${{a}_{7}}=30$, we get,

$a+6d=30..........(4)$

Subtracting equation $\left( 3 \right)$ from equation $\left( 4 \right)$, we get,

$\begin{align}

& \left( a+6d \right)-\left( a+2d \right)=30-18 \\

& \Rightarrow 4d=12 \\

& \Rightarrow d=3..........\left( 5 \right) \\

\end{align}$

Substituting $d=3$ from equation $\left( 5 \right)$ in equation $\left( 3 \right)$, we get,

$\begin{align}

& a+2\left( 3 \right)=18 \\

& \Rightarrow a+6=18 \\

& \Rightarrow a=12...........\left( 6 \right) \\

\end{align}$

In the question, we have to find the sum of $17$ terms of the A.P. Substituting $a=12,d=3,n=17$ in

equation $\left( 2 \right)$, we get,

$\begin{align}

& {{S}_{17}}=\dfrac{17}{2}\left( 2\left( 12 \right)+\left( 17-1 \right)\left( 3 \right) \right) \\

& \Rightarrow {{S}_{17}}=\dfrac{17}{2}\left( 24+\left( 16 \right)\left( 3 \right) \right) \\

& \Rightarrow {{S}_{17}}=\dfrac{17}{2}\left( 24+48 \right) \\

& \Rightarrow {{S}_{17}}=\dfrac{17}{2}\left( 72 \right) \\

& \Rightarrow {{S}_{17}}=17\times 36 \\

& \Rightarrow {{S}_{17}}=612 \\

\end{align}$

Hence, the sum of $17$ terms of this A.P. is equal to 612.

Note: There is a possibility that one may make a mistake while applying the formula of sum of an A.P. Sometimes, we apply this formula as ${{S}_{n}}=\dfrac{n}{2}\left( a+\left( n-1 \right)d \right)$ instead of ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.

difference of the AP. Using the formula of general term of an AP, we can get two equations by using

the information which is given in the question. Solve these equations to find the first term and the

common terms of the A.P.

Before proceeding in the question, we will first write down all the formulas that are

required to solve this problem.

Let us consider an arithmetic progression having its first term equal to $'a'$ and the common

difference equal to $'d'$ .

The ${{n}^{th}}$ term of this arithmetic progression (denoted by ${{a}_{n}}$) is given by the formula,

${{a}_{n}}=a+\left( n-1 \right)d..........(1)$

The sum of $n$ terms of this arithmetic progression (denoted by ${{S}_{n}}$) is given by the formula,

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

In the question, it is given that the third term i.e. ${{a}_{3}}$ of this arithmetic progression is equal to

18.

$\Rightarrow {{a}_{3}}=18$

If we substitute $n=3$ in equation $\left( 1 \right)$, we get,

${{a}_{3}}=a+\left( 3-1 \right)d$

$\Rightarrow {{a}_{3}}=a+2d$

Substituting this ${{a}_{3}}$ in the equation ${{a}_{3}}=18$, we get,

$a+2d=18..........(3)$

Also, it is given in the question that the seventh term of this arithmetic progression is equal to 30.

$\Rightarrow {{a}_{7}}=30$

If we substitute $n=7$ in equation $\left( 1 \right)$, we get,

${{a}_{7}}=a+\left( 7-1 \right)d$

$\Rightarrow {{a}_{7}}=a+6d$

Substituting this ${{a}_{7}}$ in the equation ${{a}_{7}}=30$, we get,

$a+6d=30..........(4)$

Subtracting equation $\left( 3 \right)$ from equation $\left( 4 \right)$, we get,

$\begin{align}

& \left( a+6d \right)-\left( a+2d \right)=30-18 \\

& \Rightarrow 4d=12 \\

& \Rightarrow d=3..........\left( 5 \right) \\

\end{align}$

Substituting $d=3$ from equation $\left( 5 \right)$ in equation $\left( 3 \right)$, we get,

$\begin{align}

& a+2\left( 3 \right)=18 \\

& \Rightarrow a+6=18 \\

& \Rightarrow a=12...........\left( 6 \right) \\

\end{align}$

In the question, we have to find the sum of $17$ terms of the A.P. Substituting $a=12,d=3,n=17$ in

equation $\left( 2 \right)$, we get,

$\begin{align}

& {{S}_{17}}=\dfrac{17}{2}\left( 2\left( 12 \right)+\left( 17-1 \right)\left( 3 \right) \right) \\

& \Rightarrow {{S}_{17}}=\dfrac{17}{2}\left( 24+\left( 16 \right)\left( 3 \right) \right) \\

& \Rightarrow {{S}_{17}}=\dfrac{17}{2}\left( 24+48 \right) \\

& \Rightarrow {{S}_{17}}=\dfrac{17}{2}\left( 72 \right) \\

& \Rightarrow {{S}_{17}}=17\times 36 \\

& \Rightarrow {{S}_{17}}=612 \\

\end{align}$

Hence, the sum of $17$ terms of this A.P. is equal to 612.

Note: There is a possibility that one may make a mistake while applying the formula of sum of an A.P. Sometimes, we apply this formula as ${{S}_{n}}=\dfrac{n}{2}\left( a+\left( n-1 \right)d \right)$ instead of ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.

Last updated date: 03rd Jun 2023

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