Question

Sum of
n
terms of ap is 3n^2-n find the first term and diffrence​

Answer 1

Answer:

First term(a) = 2

Common difference(d) = 16

Step-by-step explanation:

Given,

Sum of 'n' terms of an A.P = 3n² - n

To Find :-

First term(a)

Common difference (d)

How To Do :-

As they given the value of sum of 'n' terms of an A.P , we need to substitute the value that in the formula of Sum of 'n' terms of an A.P . After that we can see that the equation satisfies any value of 'n'. So to get the value of 'a'(first term) we need to take the value of 'n' as 1 because If we took value of 'n' = 1 then 'd'(common difference) becomes zero so it is easy to find the value of 'a'. After we need to take 'n' = 2 and we need to substitute in the equation and we also need to substitute the value of 'a' that we have obtained before. Then we will get the value of 'd'.

Formula Required :-

Sum of 'n' terms of an A.P :-

$S_n=\dfrac{n}{2}[2a+(n-1)d]$

Solution :-

$3n^2-n=\dfrac{n}{2}[2a+(n-1)d]$

[ ∴ S_n = n/2 [2a + (n - 1)d]

Hence we can say that the above equation can be suitable for any value of 'n'.

Taking 'n' = 1 :-

$3(1)^2-(1)=\dfrac{1}{2}[2a+(1-1)d]$

$3(1)-1=\dfrac{1}{2}[2a+(0)d]$

$3-1=\dfrac{1}{2}(2a)$

2 = a

∴ First term = a = 2

Taking 'n' = 2 :-

$3(2)^2-2=\dfrac{1}{2}[2(2)+(2-1)d]$

$3(4)-2=\dfrac{1}{2}[4 + (1)d]$

$12-2=\dfrac{1}{2}[4+d]$

$10=\dfrac{4+d}{2}$

$4+d=10\times 2$

4 + d = 20

d = 20 - 4

∴ Common difference = d = 16