Question

The sum of first m terms of an A.P. is $4m^2-m$. if it's $n^{th}$term is $107$,find the value of n.

Answer 1

(Hint: Use the formula of sum of first n terms of A.P. and find first term of A.P. with the help of sum of n terms of A.P.)
The sum of terms is given as,
$$S_m=4m^2-m...\text{(1)}$$
Let $$a_n$$ be $$n^{th}$$ the term of A.P., then we get,
$$a_1=S_1=4(1{)}^2-1=4-1=3$$
Now, we know that,
$$S_n={\displaystyle \frac{n}{2}}(a+a_n)...\text{(2)}$$
Also, the value of $$a_n$$ is given as
$$a_n=107$$
Using the equations and, we get,
$$S_n=4n^2-n={\displaystyle \frac{n}{2}}(a_1+a_n)$$
$$4n-1=\left({\displaystyle \frac{3+107}{2}}\right)$$
$$4n-1=55$$
$$n={\displaystyle \frac{56}{4}}$$
$$\Rightarrow n=14$$
So, the required solution is $$n=14$$.

Note: In order to solve these types of questions, the first term needs to be calculated first so that the formula for calculating the $$n^{th}$$term or the sum, can be applied.