Advertisements
Advertisements
प्रश्न
Find the sum of first n odd natural numbers
Advertisements
उत्तर
In this problem, we need to find the sum of first n odd natural numbers.
So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.
So here
First term (a) = 1
Common difference (d) = 2
So, let us take the number of terms as n
Now, as we know,
`S_n = n/2[2a + (n -1)d ]`
So for n terms,
`S_n = n/2 [2(1) + (n -1)2]`
`= n/2 [2 + 2n- 2]`
`= n/2 (2n)`
`= n^2`
Therefore, the sum of first n odd natural numbers is `S_n = n^2`
APPEARS IN
संबंधित प्रश्न
Find the sum of the following APs.
`1/15, 1/12, 1/10`, ......, to 11 terms.
Find the sum of first 40 positive integers divisible by 6.
Find the 12th term from the end of the following arithmetic progressions:
3, 5, 7, 9, ... 201
If numbers n – 2, 4n – 1 and 5n + 2 are in A.P., find the value of n and its next two terms.
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
Find the sum of first n even natural numbers.
Find the sum of all 2 - digit natural numbers divisible by 4.
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.
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
Q.6
Q.2
Q.18
In a Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find that:
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
Find the sum of first 10 terms of the A.P.
4 + 6 + 8 + .............
Find the value of x, when in the A.P. given below 2 + 6 + 10 + ... + x = 1800.
If a = 6 and d = 10, then find S10.
In an AP if a = 1, an = 20 and Sn = 399, then n is ______.
Find the sum of last ten terms of the AP: 8, 10, 12,.., 126.
Solve the equation
– 4 + (–1) + 2 + ... + x = 437
k + 2, 2k + 7 and 4k + 12 are the first three terms of an A.P. The first term of this A.P. is ______.
