Question

The sum of three consecutive multiples of 7 is 63. Find these multiples.

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 first term and the other two terms of the AP.

Complete Step-by-Step solution:
We know that the consecutive multiples of 7 make an AP with common difference ( d ) = 7 , first term (a), number of terms ( n ) = 3 and sum of the terms of the AP ( ${S_n}$ ) = 63
Using formula for sum of the terms of an AP
  ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
$ \Rightarrow 63 = \dfrac{3}{2}\left( {2a + \left( {3 - 1} \right)7} \right)$
$ \Rightarrow 63 \times 2 = 3\left( {2a + 14} \right)$
$ \Rightarrow \dfrac{{126}}{3} = 2a + 14$
$ \Rightarrow 42 = 2a + 14$
$ \Rightarrow a = \dfrac{{42 - 14}}{2} = \dfrac{{28}}{2} = 14$
Hence the first term is 14 .
Now ${a_n} = a + \left( {n - 1} \right)d$
Second term = 14 + 7 = 21 ( a + d )
Third term = 14 + 14 = 28 ( a + 2d )
Therefore the multiples are 14 , 21 , 28.

Note: Recall the formula of sum of an AP and also understand the concept working while generating the AP from the given information. Remember that if the first term is known then the other two following terms could be found by using the formula of ${n^{th}}$ term of an AP or by directly adding the common difference to the consecutive previous term.