Question

Which term of the A.P 3, 15, 27, 39,… will be 132 more than the 54th term?

Answer 1

Hint: Assume that the ${{m}^{th}}$ term is 132 more than the 54th term of the given A.P. Use the fact that in an A.P ${{a}_{n}}=a+\left( n-1 \right)d$, where a, n and d have their usual meanings. Put n = 54 to get the value of the 54th term. Hence find the value of the ${{m}^{th}}$ term. Use ${{a}_{m}}=a+\left( m-1 \right)d$ and hence form an equation in m. Solve for m.

Complete step-by-step answer:
Let the ${{m}^{th}}$ term of the A.P be 132 more than the 54th term.
Now, we have
a = 3 and d = 15-3=12
We know that ${{a}_{n}}=a+\left( n-1 \right)d$
Put n = 54, a = 3 and d = 12, we get
${{a}_{54}}=3+53\times 12=3+636=639$
Hence the value of ${{m}^{th}}$ term is 639+132 = 771.
Also, we have
${{a}_{n}}=a+\left( n-1 \right)d$
Put n = m, a = 3 and d = 12, we get
${{a}_{m}}=3+12\left( m-1 \right)$
Hence, we have
$3+12\left( m-1 \right)=771$
Subtracting both sides by 3, we get
$12\left( m-1 \right)=768$
Dividing both sides by 12, we get
$m-1=64$
Adding 1 on both sides, we get
$m=65$
Hence the 65th term of the A.P is 132 more than the 54th term of the A.P.

Note: Alternative solution:
Let the ${{m}^{th}}$ term be p more than ${{n}^{th}}$term of an A.P
Hence, we have
a+(m-1)d = a+(n-1)d+p
Subtracting a on both sides, we get
(m-1) d = (n-1)d+p
Dividing both sides by d, we get
$m-1=n-1+\dfrac{p}{d}$
Adding 1 on both sides, we get
$m=n+\dfrac{p}{d}$
Put n = 54, p = 132 and d= 12, we get
$m=54+\dfrac{132}{12}=54+11=65$
Hence, m = 65, which is the same as obtained above.