Question

which term of the A.P. 3, 15, 27, 39,... will be 120 more than its 21st term?

Answer 1

In A.P 3, 15, 27, 39, the $\bold{21^{st}}$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  

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

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:  

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

Answer 2

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