Question

Find the sum of first n








odd nat



ural numbers. Also find the sum of first 26terms of the AP:1,3,5,7

Answer 1

Solution:
a=1
d=2

sum of first n odd natural numbers =
=n/2(2a+(n-1)d)
=n/2(2+(n-1)2)
=n+n(n-1)
${n}^{2}$

now sum of 26 terms =
=26*26
=676
= 676
hope that's helpful
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Answer 2

The series will be
1, 3, 5, 7……(n terms)

here a= 1 d= 2 &
Using A.P, Sum of n terms= (n/2)(2a+(n-1)d)

sum of first n odd No.s = n/2(2*1+(n-1)*2)
= n/2 (2+2n-2)
= n(2n/2)
= n²

Sum of first 26 odd No.s = 26²
= 676

;)
hope it helps