Question

Which term of the A.P .21, 42, 63, 84, …..is 231?

Answer 1

Hint: We solve this question by using the concept of Arithmetic Progression A.P. We need to use the formula for the $n^{th}$ term of an A.P and this is given as ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d,$ where ${{a}_{n}}$ is the $n^{th}$ term, ${{a}_{1}}$ is the first term of the series, and d is the common difference between any two consecutive terms. Using this, we determine which term is 231 in the above sequence.

Complete step by step answer:
In order to solve this question, let us first write the equation down to calculate the $n^{th}$ term of an Arithmetic Progression. This is given by ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d,$ where ${{a}_{n}}$ is the $n^{th}$ term, ${{a}_{1}}$ is the first term of the series, and d is the common difference between any two consecutive terms.
We need to obtain the required data from the given series before substituting in the formula. The first term in the series is ${{a}_{1}}=21$ and the common difference can be calculated by subtracting any two consecutive terms as $d=42-21=21.$ It is also given that the $n^{th}$ term of the series is 231 and we need to calculate the value of n. Using these in the formula,
$\Rightarrow {{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$
Substituting the values,
$\Rightarrow 231=21+\left( n-1 \right)21$
Multiplying the terms in the right-hand side,
$\Rightarrow 231=21+21n-21$
Subtracting the terms in the right-hand side, we get
$\Rightarrow 231=21n$
Dividing both sides by 21,
$\Rightarrow \dfrac{231}{21}=n$
Simplifying,
$\Rightarrow 11=n$
Hence, 231 is the $11^{th}$ term of the A.P.

Note: We need to know the concept of Arithmetic Progression in order to solve such sums. Remembering its formula for the nth term can help us greatly or else we might have to find all the terms of the series one by one up to the $11^{th}$ term. It is important to know the concept of common difference too in the case of arithmetic progression.