Question

If $8$ times the ${{8}^{th}}$ term of an arithmetic progression is equal to $12$ times the ${{12}^{th}}$ term, then the ${{20}^{th}}$ term is
(a)$12$ times to ${{8}^{th}}$ term.
(b)$8$ times to ${{12}^{th}}$ term.
(c)0
(d)Cannot be determined.

Answer 1

Hint: To solve the question given above, we will assume that the first term of the given arithmetic progression is ‘a’ and the common difference of this arithmetic progression is ‘d’. Then we will find the ${{8}^{th}}$ and ${{12}^{th}}$ term of the given arithmetic progression and then according to question, we will apply the relation $8$ (${{8}^{th}}$ term)$=12$(${{12}^{th}}$ term). With the help of this we will find ‘a’ in terms of ‘d’. Then, we will find the ${{20}^{th}}$ term by using the formula: ${{a}_{x}}=a+\left( x-1 \right)d$.

Complete step-by-step answer:
It is given in the question that the $8$ times the ${{8}^{th}}$ term of an AP is equal to $12$ times the ${{12}^{th}}$ term of an AP. Before we solve the question we will find out what is an AP. An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. The general form of an AP is shown below:
AP: $a,a+d,a+2s,a+3d,..........$
Where ‘a’ is the first term and ‘d’ is the common difference.
Let us assume that the first term of the given AP is ‘a’ and the common difference is ‘d’. Thus the ${{n}^{th}}$ term of an AP will be given by the formula:
${{a}_{n}}=a+\left( n-1 \right)d$
Thus, the ${{8}^{th}}$ term of the AP will be:
${{a}_{8}}=a+\left( 8-1 \right)d$
$\Rightarrow {{a}_{8}}=a+7d......\left( 2 \right)$
Similarly, the ${{12}^{th}}$ term of an AP will be:
${{a}_{12}}=a+\left( 12-1 \right)d$
$\Rightarrow {{a}_{12}}=a+11d.......\left( 2 \right)$
Now, it is given in question that:
$8{{a}_{8}}=12{{a}_{12}}........\left( 3 \right)$
 Now, we will put the values of ${{a}_{8}}$ and ${{a}_{12}}$ from (1) and (2) into (3). Thus, we will get:
$8\left( a+7d \right)=12\left( a+11d \right)$
$\Rightarrow 8a+56d=12a+132d$
$\Rightarrow 12a-8a=-132d+56d$
$\Rightarrow 4a=-76d$
$\Rightarrow a=-19d.......\left( 4 \right)$
Now, we have to find the ${{20}^{th}}$ term of the AP. Thus, we have:
${{a}_{20}}=a+\left( 20-1 \right)d$
$\Rightarrow {{a}_{20}}=a+19d.......\left( 5 \right)$
Now, we will put the value of a form (4) to (5). Thus, we will get:
$\Rightarrow {{a}_{20}}=-19d+19d$
$\Rightarrow {{a}_{20}}=0$
Hence, option (c) is correct.

Note: The above question can also be solved alternatively by the following method:
$8{{a}_{8}}=12{{a}_{12}}$
Now, let us assume that ${{a}_{8}}=p$ then ${{a}_{12}}$ will be equal to $p+4d$. Thus, we have:
$8p=12\left( p+4d \right)$
$\Rightarrow 8p=12p+48d$
$\Rightarrow -4p=48d$
$\Rightarrow p=-12d$
$\Rightarrow {{a}_{8}}=-12d$
Also, ${{a}_{8}}=a+7d$
$\Rightarrow a+7d=-12d$
$\Rightarrow a+19d=0$
$\Rightarrow a+\left( 20-1 \right)d=0$
$\Rightarrow {{a}_{20}}=0$.