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 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 VerifiedHint: 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, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n- \right)d \right]$. |