Question

Which term of the AP 3 , 15 , 27 , 39 ,.. will be 120 more than its 21st term ?​

Answer 1

$In \: A.P \: 3, 15, 27, 39, \: the \: \bold{21^{st}}</p><p> \: term \: is \\ 243 \: which \: is \: 120 \: more \: than \: \\ required \: term \: 363.$

Solution:

Common difference of series = (n+1) – n

We get 15 – 3, 27 – 15

Common difference = 12

First term of series = 3.

Therefore, the 21st term of the series is

$\begin{gathered}\begin{array}{l}{T_{21}=a+d(n-1)} \\ {T_{21}=3+12(21-1)} \\ {T_{21}=243}\end{array}\end{gathered}$

Required term after adding 120 = 120 + 243 = 363.

$Let \: \bold {T_{n}=363}$

First value of the series is a = 3

Common difference = 12

$\bold{n^{th}} \: term \: is \: 363.$

Value of n:

$</p><p>\begin{gathered}\begin{array}{l}{T_{n}=3+12(n-1)} \\ {363=3+12(n-1)} \\ {n=31}\end{array}\end{gathered}$

Answer 2

Answer:

n=31is the correct answer..