An arithmetic series has nine terms ,4th term is10 and the 7th term is 16 .Find the sum of the series
d = (16-10)/(7-4) = 2
a = 10-3d = 4
S9 = 9/2 (2*4+8*2) = 108
nth term=a+(n-1)d
10=a+(4-1)d
10=a+3d
a+3d=10....eqn(i)
16=a+(7-1)d
16=a+6d
a+6d=16....eqn(ii)
a+3d=10
a+6d=16 solve using elimination method
-3d=-6
d=2
a+3×2=10
a+6=10
a=4
Sn=n/2[2a+(n-1)d]
=9/2[2×4+(9-1)2]
=9/2[8+16]
=9/2×24
=108