Question

How do you find the sum of first $30$ positive multiples of $3?$

Answer 1

Hint: To find the sum of first thirty positive multiples, first write a few initial terms to get an idea about the series whose sum you have to find. Then you will find out that the series is an arithmetic series, so simply calculate the value of common difference of the series by subtracting two consecutive terms and then use the below formula to calculate the required sum:
${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right],\;{\text{where}}\,{S_n},\;n,\;a\;{\text{and}}\;d$ are sum of terms first “n” terms, number of terms, first term of the series and its common difference respectively.

Complete step by step solution:
In order to find the sum of the first $30$ positive multiples of $3$, we will first write few of its terms to understand the series as follows
$
  3 \times 1 = 3 \\
  3 \times 2 = 6 \\
  3 \times 3 = 9 \\
  3 \times 4 = 12 \\
 $
So the given series is progressing in this way, $3,\;6,\;9,\;12,...$
We can see that it is showing the behavior of an arithmetic series
So calculating its common difference in order to find the sum of its first $30$ terms
$d = {u_{n + 1}} - {u_n} = 3 \times (n + 1) - 3n = 3n + 3 - 3n = 3$
Now, using the formula for sum of first “n” terms of arithmetic series to proceed further and calculate the required sum as follows
${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
Here we have calculated $d = 3$
And we can see that $a = 3$
And to question $n = 30$
$
  \therefore {S_n} = \dfrac{{30}}{2}\left[ {2 \times 3 + (30 - 1)3} \right] \\
   = 15\left[ {6 + 29 \times 3} \right] \\
   = 15\left[ {6 + 87} \right] \\
   = 15 \times 93 \\
   = 1395 \\
 $
$\therefore 1395$ is the required sum.

Note: When tackling these type of problems where actual series is not given but some information about the series is given then your first and compulsory step should be to write the series with its first few terms with help of given information, so that you can get an idea about the type of series and can simply solve it further.