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
•
Total views: 304.2k
•
Views today: 8.82k
Answer
304.2k+ views
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] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$
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] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE
