Question

What is the number of terms in the series 117, 120, 123, 126,….., 333?
A.72
B.73
C.76
D.79

Answer 1

Hint: Arithmetic Progression: An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. Arithmetic sequence can be finite or infinite.
As we know that the arithmetic progression will be;
 $ \Rightarrow {a_l} = {a_1} + (n - 1)d $
 $ {a_1} = first\; term $
 $ {a_l} = last\; term $
 $ d = common - difference $
 $ n = number - of - terms $

Complete step-by-step answer:
Given,
Series= 117, 120, 123, 126, …, 333
Here
 $ {a_1} = 117 $
 $ {a_2} = 120 $
 $ {a_3} = 123 $
 $ {a_l} = 333 $
 $ \Rightarrow d = {a_2} - {a_1} $
 $ \Rightarrow d = 120 - 117 $
 $ \Rightarrow d = 3 $
n=?
Using the formula of AP series.
 $ \Rightarrow {a_l} = {a_1} + (n - 1)d $
 $ \Rightarrow 333 = 117 + (n - 1)3 $
 $ \Rightarrow 333 - 117 = (n - 1)3 $
 $ \Rightarrow (n - 1)3 = 216 $
 $ \Rightarrow (n - 1) = \dfrac{{216}}{3} $
 $ \Rightarrow (n - 1) = 72 $
 $ \Rightarrow n = 72 + 1 $
 $ \Rightarrow n = 73 $
So the answer is (B) 73.
So, the correct answer is “Option B”.

Note: There are three types of progression arithmetic progression, geometric progression and harmonic progression. Arithmetic progression is based on common differences. Geometric progression is based on common ratio. And harmonic progression is based on reciprocals of arithmetic progression.