Question

The sum of first $7$ terms of an A.P. is $63$ and the sum of its next $7$ terms is \[161\] . Find the ${{28}^{th}}$ term of the A.P.

Answer 1

Hint: We need to find the ${{28}^{th}}$ term of the A.P. The sum of first 7 terms and next seven terms are given. By using the formula ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$ , we will get an equation in terms of $a$ and $d$ . The sum of next 7 terms is given. So the sum of first 14 terms can be obtained by adding the given data. Again by using the formula, ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$, we will get another equation in terms of $a$ and $d$ . By solving these two equations, we will get the values of $a$ and $d$ . Now ${{28}^{th}}$ term can be found by substituting these values in the formula ${{a}_{n}}=a+(n-1)d$ .

Complete step by step answer:
We need to find the ${{28}^{th}}$ term of the A.P.
Let us denote the sum of the first $7$ terms as ${{S}_{7}}$.
Therefore, ${{S}_{7}}=63$ .
We know that the sum of $n$ terms is given by ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$ , where $a$ is the first term of the A.P. , $n$ is the number of terms and $d$ is the common difference.
Therefore, ${{S}_{7}}=\dfrac{7}{2}\left( 2a+(7-1)d \right)=63$
Simplifying the above equation, we get
$\dfrac{7}{2}\left( 2a+6d \right)=63$
Taking 2 common from numerator of the LHS, we get
$63=\dfrac{7}{2}\times 2\left( a+3d \right)$
By simplifying the above equation we get
$a+3d=9...(i)$
Now, it is given that the sum of the next $7$ terms is $161$.
So, \[\text{sum of the first }14\text{ }\!\!~\!\!\text{ terms = sum of first }7\text{ terms + sum of next }7\text{ terms}\]
i.e ${{S}_{14}}=63+161=224$
Now using the formula, ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$ , we get
${{S}_{14}}=\dfrac{14}{2}\left( 2a+(14-1)d \right)=224$
$\Rightarrow 7\left( 2a+13d \right)=224$
Collecting constant terms on one side, we get
$2a+13d=\dfrac{224}{7}$
$2a+13d=32...(ii)$
Now let us solve the equations $(i)$ and $(ii)$ .
From $(i)$ , we can get
$a=9-3d...(iii)$
Now, substitute the value of $a$ in equation $(ii)$ .
$2(9-3d)+13d=32$
By solving, we get
$18-6d+13d=32$
$\Rightarrow 18+7d=32$
Collecting the constant terms to one side, we get
$7d=32-18$
$\Rightarrow 7d=14$
$\Rightarrow d=2$
Now substitute the value of $d$ in equation $(iii)$ , we get
$a=9-3\times 2=9-6$
$\Rightarrow a=3$
Now we have to find the ${{28}^{th}}$ term. For this, we know that ${{a}_{n}}=a+(n-1)d$ .
Let us substitute the values of $a,d$ and $n$ , where $n=28$ .
Therefore, ${{a}_{28}}=3+(28-1)2$
By solving, we will get
${{a}_{28}}=3+27\times 2$
$\Rightarrow {{a}_{28}}=3+54$
$\Rightarrow {{a}_{28}}=57$
Hence the ${{28}^{th}}$ term of the A.P. is $57$ .


Note:
 This is the normal procedure to the questions of this type. Whenever a term of an A.P. is to be found, we will be using the formula ${{a}_{n}}=a+(n-1)d$ . The values for $a$ and $d$ can be obtained from the given data. The two equations can also be solved by the elimination method.