Question

if the nth term of an AP is (2n+1), find the sum of first 'n' terms of the AP

Answer 1

Answer:

answer for the given problem is given

Answer 2

Given

• nth term of an Ap = (2n+1)

To find

• The sum of first n terms of A.P

We have,

$\sf a_n= 2n+1 \\ \\ \sf a_1= 2(1)+1=2+1= 3 \\ \\ \sf a_2= 2(2)+1= 4+1=5\\ \\ \sf a_3= 2(3)+1= 6+1= 7 \\ \\ \sf d = a_2-a_1= 5-3= 2$

Now we have to find the sum of n terms

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

We have

• Common difference (d)= 2
• First term (a)= 3

$\longrightarrow\sf S_n= \dfrac{n}{2}[2\times 3+(n-1)2]\\ \\ \longrightarrow\sf S_n= \dfrac{n}{2}(6+2n-2)\\ \\ \longrightarrow\sf S_n= \dfrac{n}{2}( 4+2n) \\ \\ \longrightarrow\sf S_n= \dfrac{n}{\cancel2}\times \cancel{2}(2+n)\\ \\ \longrightarrow\sf S_n= n(n+2)\\ \\ \longrightarrow\sf or \ S_n= n^2+2n$

$\boxed{\sf \ Sum \ of \ first \ n \ terms = n(n+2)}$