Question

If the nth term of a sequence is 8 – 5n. Show that the sequence is an AP.

Answer 1

Hint: First of all consider the nth term of the sequence and substitute the value of n = 1, 2, 3, 4, 5 in it to find the first, second, third, fourth, and fifth term of the given sequence. Now, find the difference between the consecutive terms and if they are equal, then show that the terms are in AP.

Complete step-by-step answer:

We are given that the nth term of the sequence is 8 – 5n. We have to show that sequence is an AP. Let us consider the nth term of the sequence given in the question.
${{T}_{n}}=8-5n....\left( i \right)$
Let us find the same terms of the sequence by substituting n = 1, 2, 3, 4… in equation (i).
By substituting n = 1, we get, the first term of the sequence as,

& {{T}_{1}}=8-5\left( 1 \right) \\\ & {{T}_{1}}=8-5=3....\left( ii \right) \\\ \end{aligned}$$ By substituting n = 2, we get the second term of the sequence as, $$\begin{aligned} & {{T}_{2}}=8-5\left( 2 \right) \\\ & {{T}_{2}}=8-10=-2....\left( iii \right) \\\ \end{aligned}$$ By substituting n = 3, we get the third term of the sequence as, $$\begin{aligned} & {{T}_{3}}=8-5\left( 3 \right) \\\ & {{T}_{3}}=8-15=-7....\left( iv \right) \\\ \end{aligned}$$ By substituting n = 4, we get the fourth term of the sequence as, $$\begin{aligned} & {{T}_{3}}=8-5\left( 4 \right) \\\ & {{T}_{3}}=8-20=-12....\left( v \right) \\\ \end{aligned}$$ By substituting n = 5, we get the fifth term of the sequence as, $$\begin{aligned} & {{T}_{3}}=8-5\left( 5 \right) \\\ & {{T}_{3}}=8-25=-17....\left( vi \right) \\\ \end{aligned}$$ Thus, we get the first five terms of the sequence as 3, – 2, – 7, – 12, – 17 Let us find the difference between the consecutive of the above sequence, we get, $${{T}_{2}}-{{T}_{1}}=-2-3=-5$$ $${{T}_{3}}-{{T}_{2}}=-7-\left( -2 \right)=-7+2=-5$$ $${{T}_{4}}-{{T}_{3}}=-12-\left( -7 \right)=-12+7=-5$$ $${{T}_{5}}-{{T}_{4}}=-17-\left( -12 \right)=-17+12=-5$$ Here, we can see that the common difference or the difference between any two consecutive terms is the same that is – 5 and we know that in the case of AP, the difference between two consecutive terms should be equal. Hence we can conclude that the given sequence is in AP. Note: In the above question, students can also compare the given nth term (8 – 5n) with the nth term of AP that is a + (n – 1)d and show that the common difference is – 5 as follows: We are given that, $${{T}_{n}}=8-5n....\left( i \right)$$ We have the general term of AP as $${{a}_{n}}=a+\left( n-1 \right)d$$ $${{a}_{n}}=a+dn-d$$ $${{a}_{n}}=\left( a-d \right)+{{d}{n}}....\left( ii \right)$$ By comparing $${{T}_{n}}\text{ and }{{a}_{n}}$$, we get, a – d = 8 $$-5n={{d}{n}}$$ d = – 5 Here, a is the first term and d is the common difference of AP.