Advertisements
Advertisements
Question
Find the sum of all odd natural numbers less than 50.
Advertisements
Solution
Odd natural numbers less than 50 are as follows:
1, 3, 5, 7, 9, ........, 49
Now, 3 – 1 = 2, 5 – 3 = 2 and so on.
Thus, this forms an A.P. with first term a = 1,
Common difference d = 2 and last term l = 49
Now, l = a + (n – 1)d
`=>` 49 = 1 + (n – 1) × 2
`=>` 48 = (n – 1) × 2
`=>` 24 = n – 1
`=>` n = 25
Sum of first n terms = `S = n/2 [a + 1]`
`=>` Sum of odd natural numbers less than 50
= `25/2 [1 + 49]`
= `25/2 xx 50`
= 25 × 25
= 625
APPEARS IN
RELATED QUESTIONS
The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`
The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its common difference.
What is the 5th term form the end of the AP 2, 7, 12, ..., 47?
The sum of the first n terms of an AP is `((3n^2)/2 + (5n)/2)`. Find its nth term and the 25th term.
Find the first term and common difference for the A.P.
`1/4,3/4,5/4,7/4,...`
Two A.P.’s are given 9, 7, 5, ... and 24, 21, 18, ... If nth term of both the progressions are equal then find the value of n and nth term.
Find whether 0 (zero) is a term of the A.P. 40, 37, 34, 31,... .
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
If a = 6 and d = 10, then find S10.
Jaspal Singh repays his total loan of Rs. 118000 by paying every month starting with the first instalment of Rs. 1000. If he increases the instalment by Rs. 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still have to pay after the 30th instalment?
