Question 128

If the sum of first 21 terms of an arithmetic progression is 609, then the $$11^{th}$$ term of progression is

Solution

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

609=21/2[2a+(21-1)d]

by taking 2 common and eliminate 2

609/21=a+10d i.e a11

a11=29

Share on Whatsapp Share on Whatsapp

Create a FREE account and get:

Download Maths Shortcuts PDF
Get 300+ previous papers with solutions PDF
500+ Online Tests for Free

Follow us on

Boost your Prep!

Download App