Question

Two A.P’s are given 9,7,5..... and 24,21,18..... if nth term of both the progressions are equal then find the value of n and nth term.

Answer 1

Hint: We will use the nth term formula of arithmetic progression to solve this question. The nth term of an A.P is \[{{\text{a}}_{\text{n}}}=\text{a+(n-1)d}\] and it is mentioned in the question that nth term of both the A.P’s are equal and hence we will find n.

Complete step-by-step answer:
Before proceeding with the question, we should know the concepts of arithmetic progression. An arithmetic progression is a sequence of numbers such that the difference of any two successive numbers is a constant.
First example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1
Second example, the sequence 3, 5, 7, 9, 11,... is an arithmetic progression with common difference 2.
If the initial term of an arithmetic progression is a and the common difference of successive numbers is d, then the nth term of the sequence is given by \[{{\text{a}}_{\text{n}}}=\text{a+(n-1)d}\]
Let us consider the A.P 9,7,5…… given in the question. We can see that the first term is 9 and the common difference is -2. So, nth term of first A.P is \[{{\text{a}}_{\text{n}}}=9\text{+(n-1)}\times (-\text{2})........\text{(1)}\]
Let us consider the A.P 24,21,18…… given in the question. We can see that the first term is 24 and the common difference is -3. So, the nth term of second A.P is \[{{\text{a}}_{\text{n}}}=24\text{+(n-1)}\times (-3)........\text{(2)}\]
It is mentioned in the question that nth term of both the progressions are equal so equating equation (1) and equation (2) we get,
 \[\Rightarrow 24\text{+(n-1)}\times (-3)=9\text{+(n-1)}\times (-\text{2})......\text{(3)}\]
Now simplifying equation (3) we get,
 \[\begin{align}
  & \,\Rightarrow 24-3\text{n}+3=9-2\text{n}+2 \\
 & \,\Rightarrow 3\text{n}-\text{2n = 24+3}-9-2 \\
 & \,\Rightarrow \text{n =16} \\
\end{align}\]
Now substituting n in equation (2) we get nth term,
 \[\begin{align}
  & \,\Rightarrow {{\text{a}}_{\text{n}}}=\text{a+(n-1)d} \\
 & \,\Rightarrow \,{{\text{a}}_{\text{n}}}=24\text{+(16-1)}\times (-3)=24-45=-21 \\
\end{align}\]
Hence n is 16 and nth term is -21.

Note: Remembering the nth term formula of an arithmetic progression is the key here. We can make a mistake in finding the common difference of both the A.P’s as 2 and 3 but we need to be careful to understand that here both the A.P’s are in decreasing order so common differences will be -2 and -3 respectively. We could also have substituted n in equation (1) and could have got the same answer for the nth term.