Please find the sum.

Asked by sunil_0010 | 1st Jul, 2009, 10:36: PM

Expert Answer:

the sum of first six terms of an A.P. with a as the first term and d as the common diff, is

=6/2[2a+5d]

=3[2a+5d]

=42(given)

so

2a+5d=14..(i)

next,

10th term/30th term=1/3(given0

So,

(a+9d)/(a+29d)=1/3

i.e.

2a=2d

so

a=d

substituting in(i) we get,

7d=14

a=2

so

d=2

so first erm =2

11th term=a+10d=2+20=22

Answered by  | 2nd Jul, 2009, 09:25: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.