Question

whose first term is 3 and 19th term is 129

Answer 1

→A=3

→A19=129

∴A + 18d=129

∴18d=129-a. ......Adding (-a)to both sides

D=126/18. ...multiplying both sides by 1/18

∴d=7

An =a + (n-1)d

Here an = 3 + (n-1)7

∴3+7n-7=an

An=7n-4

A1=3

A2=7(2)-4=10

A3=7(3)-4=17

A4=7(4)-4=24

And so on...

Hope this helps

Answer 2

Given a = 3  ---- (1)

Given a + 18d = 129 ---- (2)

On solving (1) and (2), we get

d = 126/18

d  = 7.

Substituted = 7 in (2), we get

a + 18(7) = 129

a + 126 = 129

a = 3.

a2 = a + d = 3 + 7 = 10

a3 = a + 2d = 3 + 17 = 17.

a4 = a + 3d = 3 + 21 = 24.


Hope this helps!