# Find the 8 th term from the end of the A.P 7, 10, 13, …….., 184.

Last updated date: 20th Mar 2023

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Hint: In the question, we are given the first term (a) and the last term (l) of the arithmetic

progression. Also, we can find the common difference (d) of this A.P. by subtracting first term from

the second term. Using the formula $l=a+\left( n-1 \right)d$, we can find the number of terms (n).

The 8 th term from the end will be the (n-8+1) th term from the start.

Before proceeding with the question, we must know all the formulas that will be required to solve

this question. Let us consider an arithmetic progression having it’s first term as ‘a’, last term as ‘l’,

common difference as ‘d’ and ‘n’ be the number of terms.

In arithmetic progressions, we have a formula,

$l=a+\left( n-1 \right)d.............\left( 1 \right)$

If we want to find r th term from the starting, it is given by the formula,

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

Also, the r th term from the end will be the (n-r+1) th $...........\left( 3 \right)$ term from the starting.

In the question, we are given an A.P. 7, 10, 13, …….., 184. The common difference (d) of this A.P. will

be the difference of the first and the second term and is equal to $10-7=3$. Let us first find the

number of terms ‘n’ of this A.P.

Substituting $a=7,l=184,d=3$ in formula $\left( 1 \right)$, we get,

$\begin{align}

& 184=7+\left( n-1 \right)\left( 3 \right) \\

& \Rightarrow \left( n-1 \right)\left( 3 \right)=177 \\

& \Rightarrow \left( n-1 \right)=59 \\

& \Rightarrow n=60 \\

\end{align}$

In the question, we are asked to find the 8 th term from the end. Using formula $\left( 3 \right)$, the

8 th term from the end will be the (60-8+1) th i.e. 53 th term from the start. Substituting $a=7,d=3,r=53$

in formula $\left( 2 \right)$, the 53 th term from the start is,

$\begin{align}

& {{a}_{53}}=7+\left( 53-1 \right)\left( 3 \right) \\

& \Rightarrow {{a}_{53}}=7+\left( 52 \right)\left( 3 \right) \\

& \Rightarrow {{a}_{53}}=7+156 \\

& \Rightarrow {{a}_{53}}=163 \\

\end{align}$

Hence, the 8 th term from the end of the A.P. is 163.

Note: There is a possibility that one may commit a mistake while converting the number of from the end to the number of terms from the starting. To convert the r th term from the end, there is a possibility that one may use the formula (n-r) instead of the formula (n-r+1) where ‘n’ is the total number of terms in the A.P.

progression. Also, we can find the common difference (d) of this A.P. by subtracting first term from

the second term. Using the formula $l=a+\left( n-1 \right)d$, we can find the number of terms (n).

The 8 th term from the end will be the (n-8+1) th term from the start.

Before proceeding with the question, we must know all the formulas that will be required to solve

this question. Let us consider an arithmetic progression having it’s first term as ‘a’, last term as ‘l’,

common difference as ‘d’ and ‘n’ be the number of terms.

In arithmetic progressions, we have a formula,

$l=a+\left( n-1 \right)d.............\left( 1 \right)$

If we want to find r th term from the starting, it is given by the formula,

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

Also, the r th term from the end will be the (n-r+1) th $...........\left( 3 \right)$ term from the starting.

In the question, we are given an A.P. 7, 10, 13, …….., 184. The common difference (d) of this A.P. will

be the difference of the first and the second term and is equal to $10-7=3$. Let us first find the

number of terms ‘n’ of this A.P.

Substituting $a=7,l=184,d=3$ in formula $\left( 1 \right)$, we get,

$\begin{align}

& 184=7+\left( n-1 \right)\left( 3 \right) \\

& \Rightarrow \left( n-1 \right)\left( 3 \right)=177 \\

& \Rightarrow \left( n-1 \right)=59 \\

& \Rightarrow n=60 \\

\end{align}$

In the question, we are asked to find the 8 th term from the end. Using formula $\left( 3 \right)$, the

8 th term from the end will be the (60-8+1) th i.e. 53 th term from the start. Substituting $a=7,d=3,r=53$

in formula $\left( 2 \right)$, the 53 th term from the start is,

$\begin{align}

& {{a}_{53}}=7+\left( 53-1 \right)\left( 3 \right) \\

& \Rightarrow {{a}_{53}}=7+\left( 52 \right)\left( 3 \right) \\

& \Rightarrow {{a}_{53}}=7+156 \\

& \Rightarrow {{a}_{53}}=163 \\

\end{align}$

Hence, the 8 th term from the end of the A.P. is 163.

Note: There is a possibility that one may commit a mistake while converting the number of from the end to the number of terms from the starting. To convert the r th term from the end, there is a possibility that one may use the formula (n-r) instead of the formula (n-r+1) where ‘n’ is the total number of terms in the A.P.

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