Question

If the sum of the first four terms of an A.P is 40 and that of the first 14 terms is 280. Find the sum of its first n terms.

Answer 1

Hint: In this question it is given that the sum of the first four terms of an A.P is 40 and that of the first 14 terms is 280. We have to find the sum of its first n terms. So to find the solution we need to find the first term and the common difference of this arithmetic progression(A.P) and after that we are able to find the summation of first n terms, which is $S_{n}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right)$.........(1)
Where a is the first term and d is the common difference.

Complete step-by-step answer:
Here given, the sum of the first four terms of an A.P is 40.
Therefore, n=4 and $S_{n}$=40.
Let us consider the first term of the A.P be a and common difference d.
Therefore by formula (1) we can write,
$S_{4}=\dfrac{4}{2} \left( 2a+\left( 4-1\right) d\right)$
$\Rightarrow 40=2\left( 2a+3d\right)$
$\Rightarrow 2\left( 2a+3d\right) =40$
$\Rightarrow \left( 2a+3d\right) =\dfrac{40}{2}$
$\Rightarrow \left( 2a+3d\right) =20$.........(2)
Also the sum of first 14 terms is 280, so by the formula (1) we can write,
${}S_{14}=\dfrac{14}{2} \left( 2a+\left( 14-1\right) d\right)$ [n=14]
$\Rightarrow 280=7\left( 2a+13d\right)$
$\Rightarrow 7\left( 2a+13d\right) =280$
$\Rightarrow \left( 2a+13d\right) =\dfrac{280}{7}$
$\Rightarrow \left( 2a+13d\right) =40$..........(3)
Now we have to solve the above equations in order to get the solution.
Subtracting (2) from (3) we get,
$\left( 2a+13d\right) -\left( 2a+3d\right) =40-20$
$\Rightarrow 2a+13d-2a-3d=20$
$\Rightarrow 10d=20$
$\Rightarrow d=\dfrac{20}{10}$
$\Rightarrow d=2$
Now by putting the value of d in equation (2) we get,
$2a+3d=20$
$\Rightarrow 2a+3\times 2=20$
$\Rightarrow 2a+6=20$
$\Rightarrow 2a=20-6$
$\Rightarrow 2a=14$
$\Rightarrow a=\dfrac{14}{2}$
$\Rightarrow a=7$
Therefore the first term of the given A.P is 7 and the common difference is 2.
Now by formula (1), the sum of first n terms,
$S_{n}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right)$
$=\dfrac{n}{2} \left( 2\times 7+\left( n-1\right) \times 2\right)$
$=\dfrac{n}{2} \times 2\left( 7+\left( n-1\right) \right)$ [ taking 2 common from each term]
$=n\left( 7+n-1\right)$
$=n\left( 6+n\right)$
$=n\left( n+6\right)$
Therefore the sum of first n terms is $n\left( n+6\right)$.

Note: While solving Arithmetic progression related problems you need to know that Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. For example, the series of natural numbers: 1,2,3,4,5,6,… is an AP, which has a common difference between two successive terms (say 1 and 2) equal to 1 (2 -1). So in general we can write the sequence of numbers as: a, (a+d), (a+2d), (a+3d),.......
Where a is the first term and d is the common difference and to find any element or summation of n terms we need to know the value of a and d.