Find the sum of the first 8 multiples of 3
The first 8 multiples of 3 add up to 108. In this article, we will see the detailed solution for this problem using the arithmetic progression formula.
Detailed Solution
- The first 8 multiples of 3 are: 3, 6, 9, 12, 15, 18, 21 and 24.
- The common difference between the consecutive multiples of 3 is 3.
- First term = a = 3
Common Difference = d = 3
Number of terms of this AP = n = 8
Last term =a_{n}=24 - Sum of an AP=\frac{n}{2}(2 a+(n-1) d)
Substituting the values of a, n and d in the above equation
Sum of the AP =\frac{8}{2}(2 \times 3+(8-1) 3)
= 4(6 + 21)
= 4 × 27
=108 - Sum of all the terms of an AP can be also written as
Sum =\frac{n}{2} (First Term + Last Term)=\frac{8}{2}(3+24)
= 4 × 27 = 108 - Therefore, when we add the first eight multiples of 3, we get the result as 108.
Frequently Asked Questions
1. Are the multiples of a number in arithmetic progression?
Ans: The series of consecutive multiples of any number is in arithmetic progression.
2. What do you mean by arithmetic progression?
Ans: The series of numbers in which the difference between any two consecutive numbers is common, then the series is known as an arithmetic progression. It is abbreviated as A.P.
3. How to calculate the sum of an A.P.?
Ans: The sum of an A.P. is given by the formula
S_{n}=\frac{n}{2}(2 a+(n-1) d)
=\frac{n}{2}(a+a+(n-1) d)
=\frac{n}{2}(First Term + Last Term)
Here,
a = first term
n = total numbers of terms
d = common difference
4. How to find the n^{t h} of an AP?
Ans: a_{n}=a+(n-1) d
Here,
a = first term
n = number of terms
d = common difference
a_{n}=n^{th}term of an AP