Question

The sum of three consecutive multiples of 5 is 45. Which is the smallest of three multiples ?
A. 10
B. 5
C. 15
D. 20

Answer 1

Hint: In this particular type of question we have to generate an AP using the information provided in the question. Then we have to apply the formula of sum of terms of an AP to get the required answer.

Complete Step-by-Step solution:
We know that the consecutive multiples of 5 make an AP with common difference ( d ) = 5 , first term or smallest term in this case (a) , number of terms ( n ) = 3 and sum of the terms of the AP ( ${S_n}$ ) = 45
Using formula for sum of the terms of an AP,
  ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
$ \Rightarrow 45 = \dfrac{3}{2}\left( {2a + \left( {3 - 1} \right)5} \right)$
$ \Rightarrow 45 \times 2 = 3\left( {2a + 10} \right)$
$ \Rightarrow \dfrac{{90}}{3} = 2a + 10$
$ \Rightarrow 30 = 2a + 10$
$ \Rightarrow a = \dfrac{{30 - 10}}{2} = \dfrac{{20}}{2} = 10$
Therefore the smallest multiple is 10.

Note: Recall the formula of sum of an AP and also understand the concept working while generating the AP from the given information . It is important to note that in such types of questions with a positive common difference the first term of the AP is the smallest term which we have to find.