Question

An A.P consists of \[57\] terms of which ${{7}^{th}}$ term is $13$ and the last term is $108$. Find the ${{45}^{th}}$ term of this A.P.

Answer 1

Hint: In order to find solution to this Arithmetic Progression Problem, we have to use a formula for finding the ${{n}^{th}}$ term of an Arithmetic Progression that is ${{a}_{n}}=a+\left( n-1 \right)d$ to find the ${{45}^{th}}$ term of this Arithmetic Series.

Complete step-by-step answer:
From our above problem, we get:
Number of terms: $n=57$
${{7}^{th}}$ term is $13$, that is we get:
${{a}_{7}}=13$
With this, we will apply ${{a}_{n}}=a+\left( n-1 \right)d$, with $n=7$ and ${{a}_{7}}=13$.
Therefore, we get:
$\Rightarrow 13=a+\left( 7-1 \right)d$
On simplifying, we get:
$\Rightarrow a+6d=13\to \left( 1 \right)$
Now, last term is $108$. Therefore, we get:
$\Rightarrow {{a}_{57}}=108$
With this again, we will apply ${{a}_{n}}=a+\left( n-1 \right)d$, with $n=57$ and ${{a}_{57}}=108$.
Therefore, we get:
$\Rightarrow 108=a+\left( 57-1 \right)d$
On simplifying, we get:
$\Rightarrow a+56d=108\to \left( 2 \right)$
Now, we will subtract equation $\left( 1 \right)$ from equation $\left( 2 \right)$.
Therefore, we get:
$\Rightarrow a+56d-\left( a+6d \right)=108-13$
On simplifying, we get:
$\Rightarrow 50d=95$
Now, on taking $50$ from LHS to RHS in denominator, we get:
$\Rightarrow d=\dfrac{95}{50}$
On simplifying, we get:
$\Rightarrow d=\dfrac{19}{10}$
As we have value of $d$, now we will substitute in equation $\left( 1 \right)$ to get value of $a$.
On substituting, we get:
$\Rightarrow a+6\left( \dfrac{19}{10} \right)=13$
On simplifying, we get:
$\Rightarrow a+\dfrac{114}{10}=13$
On simplifying, we get:
$\Rightarrow a+\dfrac{57}{5}=13$
Now, on taking $\dfrac{57}{5}$ on RHS we get:
$\Rightarrow a=13-\dfrac{57}{5}$
On simplifying, we get:
$\Rightarrow a=\dfrac{65-57}{5}$
On simplifying, we get value of $a$ as:
$\Rightarrow a=\dfrac{8}{5}$
Now, as we have all terms to find the ${{45}^{th}}$ term, we will substitute it in our formula ${{a}_{n}}=a+\left( n-1 \right)d$ to get the answer.
With $n=45$, $a=\dfrac{8}{5}$, $d=\dfrac{19}{10}$
On substituting, we get:
${{a}_{45}}=\dfrac{8}{5}+\left( 45-1 \right)\times \dfrac{19}{10}$
On simplifying our equation, we get:
${{a}_{45}}=\dfrac{8}{5}+44\times \dfrac{19}{10}$
On simplifying, we get:
${{a}_{45}}=\dfrac{8}{5}+\dfrac{418}{5}$
On further simplification, we get:
${{a}_{45}}=\dfrac{426}{5}$
On simplifying, we get ${{45}^{th}}$ term as :
${{a}_{45}}=85.2$
Therefore, ${{45}^{th}}$ term of our A.P. is $85.2$.

Note: Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
We have two major formulas which is related to ${{n}^{th}}$ term of Arithmetic Progression:
To find the ${{n}^{th}}$ term of A.P: ${{a}_{n}}=a+\left( n-1 \right)d$
To find sum of ${{n}^{th}}$ term of A.P: $S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Based on given question of an Arithmetic Progression, we have to decide which formula we have to use.