Question

How many terms are there in the series :

(i) 4, 7 , 10, 13, ................ , 148 ?

(ii) 0.5, 0.53, 0.56, .................... , 1.1 ?

(iii) $\frac{3}{4},1,1\frac{1}{4},....................,3?$

Answer 1

(i)common difference (d)

A + (n-1)d = last term
4 + (n-1)3 = 148
(n-1)3 = 148 - 4
(n-1) = 144/3
n = 48 + 1
n = 49



(ii)common difference =0.03

A + (n-1)d = last term
0.5+(n-1)0.03=1.1
n-1=0.6/0.03
n-1=20
n=21

(iii)