Question

The third term of an Ap is 8 and the ninth term exceeds three times the third term by 2 find the sum of its first 19 terms

Answer 1

Answer:

Step-by-step explanation:

T(3)=8 - - - - - - - (1)

& T(9)=3*T(3)+2 - - - - - - - - - (2)

Find S(19)

T(n) = a+(n-1) d

Hence equation (1) becomes,

a+(3-1)d=8

a+2d=8

a=8-2d - - - - - - - - - - (3)

Also equation (2) becomes,

a+(9-1) d = 3×[a+(3-1) d] +2

a+8d=[3*(a+2d)]+2

a+8d=(3a+6d)+2

a+8d=3a+6d+2

Rearranging,

8d-6d=3a-a+2

2d=2a+2

Dividing by 2 on both sides,

d=a+1 - - - - - - - - - - - - - - - - - - - (4)

Substituting (3) in (4) we get,

d=(8-2d)+1

d=8-2d+1

d=9-2d

3d=9

d=9/3

d=3

Substituting value of d in (3)

a=8-2(3)

a=8-6

a=2

Hence a=2 & d=3

We know that, S(n) = n/2 × [2a+(n-1) d]

S(19) = 19/2 × [2*2+(19-1)*3]

= 19/2 * [4+18*3]

= 19/2 * [4+54]

=19/2 * 58

=19*29

=551

Sum of first 19 terms of grid AP is 551

Read more on Brainly.in - https://brainly.in/question/2434804#readmore