Question

in an Ap if an = 2n+1 then sn=??​

Answer 1

An=2n+1

A1=3

A2=5

A3=7

a=3

d=5-3=2

Sn=n/2(2a+(n-1)d)

=n/2(2×3+2n-2)

=n/2(6-2+2n)

=n/2(4+2n)

=n²+2n

Answer 2

sn = n(n+2)

• $a_{n} = 2n+1$ is given.
• Now the first term, $a_{1$ = 2+1 = 3 and second term, $a_{2}$ = 4+1 = 5
• So common difference,d = $a_{2} -a_{1} = 5-3=2$
• Hence sum of the series , $s_{n} = \frac{n}{2} [2a_{1} +(n-1)d]$
• Now putting the values we get $s_{n} = \frac{n}{2} [2*3 +(n-1)*2] = \frac{n}{2} [6 +2n-2] = \frac{n}{2} [2n+4] = n(n+2)$
• So the required sn = n(n+2)