Advertisements
Advertisements
Question
If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is
Options
3200
1600
200
2800
Solution
In the given problem, we need to find the sum of 40 terms of an arithmetic progression, where we are given the first term and the common difference. So, here we use the following formula for the sum of n terms of an A.P.
`S_n = n/2 [2a + (n-1) d]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
Given,
First term (a) = 2
Common difference (d) = 4
Number of terms (n) = 40
So, using the formula we get,
`S_40 = 40/2 [2(2) + (40 - 1)(4)]`
=(20) [4 + (39)(4)]
=(20)[4 + 156]
=(20)(160)
= 3200
Therefore, the sum of first 40 terms for the given A.P. is `S_40 = 3200`.
APPEARS IN
RELATED QUESTIONS
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Find the sum of first 15 multiples of 8.
If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero
The next term of the A.P. \[\sqrt{7}, \sqrt{28}, \sqrt{63}\] is ______.
If the sum of n terms of an A.P. is 3n2 + 5n then which of its terms is 164?
If Sn denote the sum of n terms of an A.P. with first term a and common difference dsuch that \[\frac{Sx}{Skx}\] is independent of x, then
Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is `(1)/(3)`. Calculate the first and the thirteenth term.
In an A.P., the sum of its first n terms is 6n – n². Find is 25th term.
The sum of first ten natural number is ______.