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

Answer

Verified

384.9k+ views

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

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which one of the following places is unlikely to be class 8 physics CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Elucidate the structure of fructose class 12 chemistry CBSE

What is pollution? How many types of pollution? Define it