Question

Which term of the sequence 72, 70, 68, 66. . . . . is 40?

Answer 1

Hint: In this question, we are given a sequence 72, 70, 68, 66 . . . . . . . We need to find the term of sequence for which the value will be 40. For this, we will first prove that the given sequence is arithmetic progression by calculating the difference between the second term and the first term, between the third term and the second term. If the difference is the same, then the given sequence will be an AP. Then we will apply the formula ${{a}_{n}}=a+\left( n-1 \right)d$ where ${{a}_{n}}$ is the ${{n}^{th}}$ term, a is the first term and d is the common difference of an AP.

Complete step-by-step solution
Here, we are given the sequence as 72, 70, 68, 66 . . . . . . . . . .
Let us check if the given sequence is an arithmetic progression or not. For this, let us make the difference between the first and second term, we get 70-72 = -2.
The difference between the second and third terms is 68-70 = -2.
The difference between the third and fourth term is 66-68 = -2.
Since all the differences are the same, the given sequence is an arithmetic progression with common difference -2. So, d = -2.
Also, we can see that the first term of the AP is 72. So, a = 72.
Now, we need to find the term whose value is 40 which means we need to find n if ${{a}_{n}}=40$.
As we know, that in an AP, ${{a}_{n}}=a+\left( n-1 \right)d$.
So putting in the values we get:
$40=72+\left( n-1 \right)\left( -2 \right)\Rightarrow 40=72-2n+2\Rightarrow 2n=34\Rightarrow n=17$.
Here ${{17}^{th}}$ term of the given AP is 40.

Note: Students should take care while calculating the common difference. They should note that common differences can be negative too. Students can get confused between the terms n and ${{a}_{n}}$. Here ${{a}_{n}}$ represents the term of the AP at ${{n}^{th}}$ position. Make sure to put 40 as the value of ${{a}_{n}}$ and not of n. While checking for AP, make sure that the difference between all consecutive terms is the same.