Question

Find the 29th term of the following sequence 5,8,11,14,17

Answer 1

Answer:

$here \: a = 5 \\ and \: d = 3 \\ so \: a {}^{29 } = a + (n - 1)d \\ = 5 + (29 - 1)3 \\ = 89$

hope it will help you

Answer 2

Given,

sequence - 5,8,11,14,17

common difference=a2-a1=8-5=3

common difference=a3-a2=11-8=3

similarly a4-a3=3 and a5-a4=3

as common difference is same this sequence is in A.P

in given A.P(Arithmetic progression)

first term=a=5

common difference=d=3

29th term=a29=a+(29-1)d {as an term =a+(n-1)d

29th term=a29=5+(28)3

29th term=a29=5+84

29th term=a29=89.

so 29th term in given sequence is 89.