Question

Find the first six terms of series in AP of which the sum to n terms is n/2(7n-1).

Answer 1

Answer:

Given Sn=n/2 (7n-1)

S1= 1/2 (7(1) -1) =6/2 =3

therefore a=3

S2= 2/2 ( 7 (2)-1) = 1 (14-1) =13

S2= 13

a+(a+d)= 13

2a+d=13 [a=3]

2 (3)+d=13

6+d=13

d=13-6

d=7

t1= a= 3

t2= a+d= 3+7 =10

t3= a+2d= 3+2 (7) = 3+14 =17

t4= a+3d= 3+3 (7)= 3+21= 24

t5= a+4d= 3+4 (7)= 3+28=31

t6= a+5d= 3+5 (7)=3+35=38

The first six terms are 3, 10, 17, 24, 31, 38