Question

Which term of the AP 8, 14, 20, 26 ... will be 72 more than its $41^{\text{st}}$ term?

Answer 1

The given AP is 8, 14, 20, 26 ... $\Rightarrow a = 8$, $d = {a_2} - {a_1} = 14 - 8 = 6$ . Also, it is asked to find the term which will be 72 more than the $41^{\text{st}}$ term.
Let n be the term which satisfies the above statement.
Then, find the value of the $n^{\text{th}}$ term and $41^{\text{st}}$ term by using the formula ${T_n} = a + \left( {n - 1} \right)d$.
According to the given condition, ${T_n} = {T_{41}} + 72$ .
Thus, find n from the above expression.

Complete step by step solution:
The given AP is 8, 14, 20, 26 ...
So, $a = 8$, $d = {a_2} - {a_1} = 14 - 8 = 6$.
Now, we are asked to find the term which will be 72 more than the $41^{\text{st}}$ term.
Let, n be the term which satisfies the above statement.
Thus, the value of $n^{\text{th}}$ term can be given by the formula ${T_n} = a + \left( {n - 1} \right)d$
$\Rightarrow {T_n} = 8 + \left( {n - 1} \right)6 \\\ \Rightarrow {T_n} = 8 + 6n - 6 \\\ \Rightarrow {T_n} = 2 + 6n$
Also, $41^{\text{st}}$ term can be given by
${T_{41}} = 8 + \left( {41 - 1} \right)6 \\\ = 8 + 40 \times 6 \\\ = 8+240 \\\ = 248$
Thus, ${T_{41}} = 248$.
As the condition is given that $n^{\text{th}}$ term of the given AP exceeds 72 by the $41^{\text{st}}$ term.
$\Rightarrow {T_n} = {T_{41}} + 72 \\\ \Rightarrow 2 + 6n = 248 + 72 \\\ \Rightarrow 6n = 320 - 2 \\\ \Rightarrow 6n = 318 \\\ \Rightarrow n = \dfrac{{318}}{6} \\\ \Rightarrow n = 53$

Thus, the required term is the $53^{\text{rd}}$ term.

Note:
Arithmetic progression (AP):
A sequence of numbers in order in which the difference between any two consecutive terms is always a constant value is called an arithmetic progression. Also, the first term of the AP must be a defined value.
For example, the series of whole numbers is an arithmetic progression as the first term of that series is 0 and the common difference is 1.