The multiples of 3 are 3,6,9,12,15,18,21,24,27,30,.... The multiples of 5 are 5,10,15,20,25,30,35,40,45,.... The intersection of these two sequences is 15,30,45,... The sum of the first numbers 1+2+3+4+...+n is n(n+1)/2. The sum of the first few multiples of k, say k+2k+3k+4k+...+nk must be kn(n+1)/2. Now you can just put these ingredients together to solve the problem. Since we are asked to look for numbers below 1000, we shall look at numbers up to the number 999. To find n, use 999/3 = 333 + remainder, 999/5 = 199 + remainder, 999/15 = 66 + remainder, by using a*(m*(m+1)/2) , where m=n/a. here a is 3 or 5 or 15, and n is 999 or 1000 but 999 is best, and then sum multiples of 3: $3((333)*(333+1)/2) = 166833$ plus multiples of 5: $5((199)*(199+1)/2) = 99500$; and subtract multiples of 15 $15((66)+(66+1)/2 )= 33165$ to get 233168. Answer  Hint: To solve this question, we need to have the knowledge of arithmetic progression. For example, let us consider an arithmetic progression a, a+d, a+2d, …….., a+(n-1)d having n terms in AP, then the sum of these n terms can be represented as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n- \right)d \right]$. Complete step-by-step answer: In this question, we are asked to find out the sum of the first five multiples of 3. We know that the first five multiples of 3 are 3, 6, 9, 12 and 15, which forms an arithmetic progression with 3 as the first term of the AP and (6 – 3) = 3 as the common difference of the AP. So, to find the first five terms of an AP, we need to know that, sum of n terms of an AP can be written as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n- \right)d \right]$. Where $n$ represents the number of terms in the AP, $a$ represents the first term of the AP and $d$ represents the common difference of the AP. So, from the question, we can write that, $n=5,a=3$ and $d=3$. Therefore, the sum of the first five multiples of 3 can be written as: $\begin{align} & {{S}_{5}}=\dfrac{5}{2}\left[ 2\left( 3 \right)+\left( 5-1 \right)\left( 3 \right) \right] \\ & \Rightarrow \dfrac{5}{2}\left[ 6+4\left( 3 \right) \right] \\ & \Rightarrow \dfrac{5}{2}\left[ 6+12 \right] \\ & \Rightarrow \dfrac{5}{2}\left[ 18 \right] \\ & \Rightarrow 5\times 9 \\ & \Rightarrow 45 \\ \end{align}$ Therefore, we can write that the sum of the first five multiples of 3 is 45.Hence, option (A) is the correct option.Note: In this question, we can also find the sum of the first five multiples of 3 by simply adding them. But if we get a larger number like 12, then the solution will get lengthier and it will be very time consuming. So, it is better to apply the formula of finding the sum of $n$ terms of AP, that is, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n- \right)d \right]$. |
What is the sum of the first three multiples of 5
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