Question

Find the sum of first 8 multiples of 3

Answer 1

Hint: First write the series of first 8 multiple of 3 then find the value of first term and the common difference of the AP. By using the sum of n term formulas of an AP series find the sum of the series by taking n is equal to 8.

Complete Step-by-step solution
Given first 8 multiples of 3-
So the series will be like this
3,6,9,.....upto 8 terms
The above series is in A.P.
Because the common difference is same
So,
First term (a) = 3
Common difference (d) = 3
No. of terms (n) = 8
Sum of terms (Sn​) = ?
As we know that, in an A.P.,
∴$${S_{8}} = \dfrac{{8}}{2}\left[ {2 \times 3 + \left( {8 - 1} \right) \times 3} \right]$$
$${S_{8}} = 4 \times \left( {6 + 21} \right)$$
$${S_8} = 4 \times 27$$
$${S_8} = 108$$

Hence the sum of the first 8 multiples of 3 is 108.

Additional information:
Sum of n terms in a sequence can be evaluated only if we know the type of sequence it is. Usually, we consider arithmetic progression, while calculating the sum of n number of terms. In this progression, the common difference between each succeeding term and each preceding term is constant. An example of AP is natural numbers, where the common difference is 1. Therefore, to find the sum of natural numbers, we need to know the formula to find it. Let us discuss here.

Sum of n terms in AP	$$\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$
Sum of n natural numbers	$$\dfrac{{{\bf{n}}\left( {{\bf{n}} + {\bf{1}}} \right)}}{{\bf{2}}}$$
Sum of square of ‘n’ natural numbers	$$\dfrac{{\left[ {{\bf{n}}\left( {{\bf{n}} + {\bf{1}}} \right)\left( {{\bf{2n}} + {\bf{1}}} \right)} \right]}}{{\bf{6}}}$$.
Sum of Cube of ‘n’ natural numbers	$${\left[ {\dfrac{{{\bf{n}}\left( {{\bf{n}} + {\bf{1}}} \right)}}{{\bf{2}}}} \right]^2}$$

Note:Sum of an AP also calculated by the formula $${S_n} = \dfrac{n}{2}\left[ {a + l} \right]$$where l is the last term of the AP series.