Question

An A.P. consists of 50 terms of which 3rd term is 12 nd the last term is 106. fnd the 29th term.

Answer 1

Answer:

Step-by-step explanation:

n=50

t3=12 =

a+2d=12  -(1)

t50=106=

a+49d=106 -(2)

Substracting 1 and 2,

a+49d - (a+2d)= 106-12

47d=94

d=94\47

d=2

a+2d=12

a=12-4

a=8

t29=?

a+28d=8+ 28d= 64