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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.

Answer Verified Verified
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 $m=n+\dfrac{p}{d}$.

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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