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Question: The sum of first m terms of an A.P. is \(4{m^2} - m\). if it's \({n^{th{\text{ }}}}\)term is \(107\)...

The sum of first m terms of an A.P. is 4m2m4{m^2} - m. if it's nth {n^{th{\text{ }}}}term is 107107,find the value of n.

Explanation

Solution

(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,
Sm=4m2m ...(1){S_m} = 4{m^2} - m{\text{ }}...{\text{(1)}}
Let an{a_n} be nth{n^{th}} the term of A.P., then we get,
a1=S1=4(1)21=41=3{a_1} = {S_1} = 4{(1)^2} - 1 = 4 - 1 = 3
Now, we know that,
Sn=n2(a+an) ...(2){S_n} = \dfrac{n}{2}(a + {a_n}){\text{ }}...{\text{(2)}}
Also, the value of an{a_n} is given as
an=107{a_n} = 107
Using the equations and, we get,
Sn=4n2n=n2(a1+an){S_n} = 4{n^2} - n = \dfrac{n}{2}({a_1} + {a_n})
4n1=(3+1072)4n - 1 = \left( {\dfrac{{3 + 107}}{2}} \right)
4n1=554n - 1 = 55
n=564n = \dfrac{{56}}{4}
n=14\Rightarrow n = 14
So, the required solution is n=14n = 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 nth{n^{th}}term or the sum, can be applied.