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

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.

Explanation

Solution

Hint: We will use the nth term formula of arithmetic progression to solve this question. The nth term of an A.P is an=a+(n-1)d{{\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 an=a+(n-1)d{{\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 an=9+(n-1)×(2)........(1){{\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 an=24+(n-1)×(3)........(2){{\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,
24+(n-1)×(3)=9+(n-1)×(2)......(3)\Rightarrow 24\text{+(n-1)}\times (-3)=9\text{+(n-1)}\times (-\text{2})......\text{(3)}
Now simplifying equation (3) we get,

& \,\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{aligned}$$ Now substituting n in equation (2) we get nth term, $$\begin{aligned} & \,\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{aligned}$$ 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.