Question

Which term of the AP : 3, 15, 27, 39, ……… will be 132 more than its 54^th term?

Answer 1

Given A.P. is \(3, 15, 27, 39, …….. \)
\(a = 3\) and  \(d = a_2 − a_1 = 15 − 3 = 12\)
\(a_{54} = a + (54 − 1) d\)
\(a_{54} = 3 + (53) (12)\)
\(a_{54}= 3 + 636 = 639\)
Now \(a_{54} +132 = 639 + 132 = 771\)
We have to find the term of this A.P. which is \(771\).
Let n^th term be \(771\).
\(a_n = a + (n − 1) d\)
\(771 = 3 + (n − 1) 12\)
\(768 = (n − 1) 12\)
\(n − 1 = \frac {768}{12}\)
\(n − 1 = 64\)
\(n = 65\)

Therefore, 65^th term was \(132\) more than 54^th term.

Alternatively,
Let n^th term be \(132\) more than 54^th term.
\(n = 54+\frac {132}{12}\)
\(n = 54+11\)
\(n = 65 ^{th}\)term.