Aptitude Discussion

Q. |
The average of first five multiples of 3 is: |

✖ A. |
8 |

✔ B. |
9 |

✖ C. |
10 |

✖ D. |
11 |

**Solution:**

Option(**B**) is correct

Basic Formula: $1, 2, 3 ... n$

If $n$ is odd, the formula is $\left(\dfrac{n+1}{2}\right)^{th}$ term

The five multiples of 3 is $3, 6, 9, 12, 15$

$\left(\dfrac{n+1}{2}\right) \Rightarrow \left(\dfrac{5+1}{2}\right)^{th}$ term

$\Rightarrow \left(\dfrac{6}{2}\right)^{th}$ term = $3^{rd}$ term

Here $3^{rd}$ term is **9**

**Edit:** For generalisation of the problem, check comment by **Sravan Reddy.**

**Shubham Palande**

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**Mayank**

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There is simple formula for this if the difference is same you have to add last term with first term and divide it by 2.

3 6 9 12 15 our 5 multiple of 3 and difference between them is 3 therefore 15+3 / 2=9

**Sravan Reddy**

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If $n$ is odd -> middle number $= {\dfrac{n+1}{2}}^{\text{th}}$ term is the answer

If $n$ is even -> average of middle two numbers $\dfrac{n}{2}$ term and $\dfrac{n}{2}+1$ term

sir, is zero not a multiple of 3?i think is it. and according to my knowledge that 5 multiples are 0,1,2,3,4 ..

waiting for your reply

**Runa**

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Very simple...The five multiples of 3 are 3, 6, 9, 12, 15 whose sum is 45.If we divide the sum by 5, we will get the average of 9.

**Ambuj**

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I still cant get the new technique..I am stuck wid the old one..

If you know how to calculate the 'Median' you'll probably able to related with the given solution in a better manner.

**Lokesh**

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Oh wow, loved the new technique to solve the question.

I tried the old method of writing all the numbers and dividing by the total number of terms.

Although both methods give the same ans, its always good to know an alternative

what if the n is an even number,

alternate is given for odd number but not for an even number.

we can also do it as

3+6+9+12+15=45

Therefore 45/5=9