     Question Answers

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

View Notes
Table of 39 - Multiplication Table of 39  Arithmetic Progression  Table of 27 - Multiplication Table of 27  Factors of 27  Table of 15  Multiples of 15  Factors of 15  Value of Tan 15  Value of Sin 15  DK Goel Solutions Class 11 Accountancy Chapter 27  