Question

prove that the sum of nth term from the begining and nth term from the end of an A.P. is constant​

Answer 1

Answer:

Step-by-step explanation:

Suppose an ap

As:- 123456.......10

A first term=1

D comman difference =1

Number of terms =11

So formula is (n/2)*2a+(n-1)d=11/2*(2+10)

11/2*12========1st eq

As:- 10987645321

A first term=10

D comman difference =-1

Number of terms =11

So formula is (n/2)*2a+(n-1)d

11/2*(20+[11-1]*-1)

11/2*(12)

So both are equal