Question

An



a.P.,has 2n+1 terms.Let so represent the sum of odd terms and se represent the sum of its even terms.Prove that so-se=a+nd,where a is the first term and d us the common difference.

Answer 1

Answer:

Step-by-step explanation:

: (n+1):n

Let ap be the pth term of the given A.P

∴ap=a+(p−1)d

S1=Sum of odd terms

(ie) S1=a1+a3+a5+......a2n+1

=n+12[a1+a2n+1]

=n+12[2a+(2n+1−1)d]

⇒S1=(n+1)(a+nd)

S2=sum of even terms

S2=a2+a4+a6+....a2n

=n2[a2+a2n]

=n2[(a+d)+(a+(2n−1)d)]

=n[a+nd]

∴S1:S2=(n+1)(a+nd):n(a+nd)

=(n+1):n