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Question
Find the sum of all odd natural numbers less than 50.
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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
RELATED QUESTIONS
If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their mth terms.
In an AP given a = 3, n = 8, Sn = 192, find d.
Find the sum of the following arithmetic progressions:
a + b, a − b, a − 3b, ... to 22 terms
In an A.P. the first term is 25, nth term is –17 and the sum of n terms is 132. Find n and the common difference.
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P.
Write the nth term of an A.P. the sum of whose n terms is Sn.
For what value of p are 2p + 1, 13, 5p − 3 are three consecutive terms of an A.P.?
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60...... is 300?
Hence find the sum of all the terms of the Arithmetic Progression (A.P.)
