Question

The 14th term of an A.P. is twice its 8th term. Its 6th term is -8, then find the sum of its first 20 terms.

Answer 1

a+ 13 d = 2(a+7d)

a + 13d = 2a + 14d

​-a -d = 0

a+ d = 0 = a2 ---- --------> 1
a + 5d = -8 -----> 2
 
2-1

a + 5d = -8
a +. d = 0
------------------
a = 2
D = -2


a 20 = 2+ 19*-2
     
      = -36

S20 = 10(2+(-36))
     
= -340

Hope it's helpful.

All the best.

Answer 2

Solution :

Given a14=2a8;a6=−8

an=a+(n−1)d

∴a+13d=2(a+2d)

a+13d=2a+14d

a=−d----------(1)

a+rd=−8

substituting for d we get,

−d+5d=−8

4d=−8

d=−2

∴a=2

sn=n2[2a+(n−1)d

S2n=2n2[2(2)+(19)(−2)]

=10[4−38]

=−340