Question

( n^{k h} operatorname{tes} m=2 n-1 )
at ( phi n=2 Rightarrow ) & ( 2 n-1=2(1)-1=1=a )
( begin{aligned} l operatorname{ten} n=2 n-1 & operatorname{sum} operatorname{ton} operatorname{ten} v &=frac{n}{2}(a+b) &=frac{n}{2}(y+2 n-1) &=n^{2} end{aligned} )

# If the nth term of an A.P. be (2n-1), then the sum of its first n terms will be (a) n2 - 1 (b) (n-1)2 + (2n-1) © (n-1)² – (2n-1) (d) a

Solution