# Which term of the following A.P

\[3,\text{ }15,\text{ }27,\text{ }39,\text{ }\ldots \]

will be 132 more than the 54th term of the A.P.

Last updated date: 22nd Mar 2023

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Answer

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Hint: Use the formula for the nth term of A.P ${{a}_{n}}=a+(n-1)d$. Where ‘a’ is the first term and ‘d’ is the common difference to find the 54th term. Then use ${{a}_{n}}=a+(n-1)d$again to find the term which is 132 more than the 54th term. Alternatively, you can use the property that if mth term is p more than nth term then $m=n+\dfrac{p}{d}$.

We know that ${{a}_{n}}=a+(n-1)d\text{ (i)}$

In the given A.P we have a = 3

d = 15-3 = 12

Put n = 54, a = 3 and d = 12 in equation (i), we get

$\begin{align}

& {{a}_{54}}=3+\left( 54-1 \right)\times 12 \\

& \Rightarrow {{a}_{54}}=3+53\times 12 \\

& \Rightarrow {{a}_{54}}=3+636 \\

& \Rightarrow {{a}_{54}}=639 \\

\end{align}$

Let the kth term of A.P be 132 more than 54th term

Hence ${{a}_{k}}={{a}_{54}}+132$

$\Rightarrow a+\left( k-1 \right)d=639+132$

Put a = 3 and d = 12

$\Rightarrow 3+\left( k-1 \right)12=771$

Subtracting 3 from both sides we get

$\begin{align}

& \Rightarrow 3+\left( k-1 \right)12-3=771-3 \\

& \Rightarrow \left( k-1 \right)12=768 \\

\end{align}$

Dividing both sides by 12, we get

$\begin{align}

& \Rightarrow \dfrac{\left( k-1 \right)12}{12}=\dfrac{768}{12} \\

& \Rightarrow k-1=64 \\

\end{align}$

Transposing 1 to RHS we get

k = 64+1 = 65

Hence the 65th term is 132 more than 54th term

Note: [a] Alternatively, let mth term is p more than nth term

Then we have

$\begin{align}

& {{a}_{m}}={{a}_{n}}+p \\

& \Rightarrow a+\left( m-1 \right)d=a+\left( n-1 \right)d+p \\

\end{align}$

Subtracting a from both sides we get

$\begin{align}

& \Rightarrow a+\left( m-1 \right)d-a=a+\left( n-1 \right)d+p-a \\

& \Rightarrow \left( m-1 \right)d=\left( n-1 \right)d+p \\

\end{align}$

Dividing by d on both sides we get

$\begin{align}

& \dfrac{\left( m-1 \right)d}{d}=\dfrac{\left( n-1 \right)d+p}{d} \\

& \Rightarrow m-1=\dfrac{\left( n-1 \right)d+p}{d} \\

\end{align}$

Transposing 1 to RHS we get

$\begin{align}

& \Rightarrow m=\dfrac{\left( n-1 \right)d+p}{d}+1 \\

& \Rightarrow m=\dfrac{\left( n-1 \right)d}{d}+\dfrac{p}{d}+1 \\

& \Rightarrow m=n-1+\dfrac{p}{d}+1 \\

& \Rightarrow m=n+\dfrac{p}{d} \\

\end{align}$

Put n = 54, p = 132 and d = 12 we get

$m=54+\dfrac{132}{12}$

m = 54+11 = 65

[b] The value of the first term was not needed in solving the question through the second method.

[c] Some of the most important formulae in A.P are:

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

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

[3] ${{S}_{n}}=\dfrac{n}{2}\left[ 2l-\left( n-1 \right)d \right]$ where l is the last term.

[4] ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$

[5] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$

__Complete step-by-step solution -__We know that ${{a}_{n}}=a+(n-1)d\text{ (i)}$

In the given A.P we have a = 3

d = 15-3 = 12

Put n = 54, a = 3 and d = 12 in equation (i), we get

$\begin{align}

& {{a}_{54}}=3+\left( 54-1 \right)\times 12 \\

& \Rightarrow {{a}_{54}}=3+53\times 12 \\

& \Rightarrow {{a}_{54}}=3+636 \\

& \Rightarrow {{a}_{54}}=639 \\

\end{align}$

Let the kth term of A.P be 132 more than 54th term

Hence ${{a}_{k}}={{a}_{54}}+132$

$\Rightarrow a+\left( k-1 \right)d=639+132$

Put a = 3 and d = 12

$\Rightarrow 3+\left( k-1 \right)12=771$

Subtracting 3 from both sides we get

$\begin{align}

& \Rightarrow 3+\left( k-1 \right)12-3=771-3 \\

& \Rightarrow \left( k-1 \right)12=768 \\

\end{align}$

Dividing both sides by 12, we get

$\begin{align}

& \Rightarrow \dfrac{\left( k-1 \right)12}{12}=\dfrac{768}{12} \\

& \Rightarrow k-1=64 \\

\end{align}$

Transposing 1 to RHS we get

k = 64+1 = 65

Hence the 65th term is 132 more than 54th term

Note: [a] Alternatively, let mth term is p more than nth term

Then we have

$\begin{align}

& {{a}_{m}}={{a}_{n}}+p \\

& \Rightarrow a+\left( m-1 \right)d=a+\left( n-1 \right)d+p \\

\end{align}$

Subtracting a from both sides we get

$\begin{align}

& \Rightarrow a+\left( m-1 \right)d-a=a+\left( n-1 \right)d+p-a \\

& \Rightarrow \left( m-1 \right)d=\left( n-1 \right)d+p \\

\end{align}$

Dividing by d on both sides we get

$\begin{align}

& \dfrac{\left( m-1 \right)d}{d}=\dfrac{\left( n-1 \right)d+p}{d} \\

& \Rightarrow m-1=\dfrac{\left( n-1 \right)d+p}{d} \\

\end{align}$

Transposing 1 to RHS we get

$\begin{align}

& \Rightarrow m=\dfrac{\left( n-1 \right)d+p}{d}+1 \\

& \Rightarrow m=\dfrac{\left( n-1 \right)d}{d}+\dfrac{p}{d}+1 \\

& \Rightarrow m=n-1+\dfrac{p}{d}+1 \\

& \Rightarrow m=n+\dfrac{p}{d} \\

\end{align}$

Put n = 54, p = 132 and d = 12 we get

$m=54+\dfrac{132}{12}$

m = 54+11 = 65

[b] The value of the first term was not needed in solving the question through the second method.

[c] Some of the most important formulae in A.P are:

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

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

[3] ${{S}_{n}}=\dfrac{n}{2}\left[ 2l-\left( n-1 \right)d \right]$ where l is the last term.

[4] ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$

[5] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$

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