Advertisements
Advertisements
प्रश्न
Find the sum of odd natural numbers from 1 to 101.
Advertisements
उत्तर
The odd natural numbers from 1 to 101 are
1, 3, 5,....., 101
The above sequence is an A.P.
∴ a = 1, d = 3 – 1 = 2
Let the number of terms in the A.P. be n.
Then, tn = 101
Since tn = a + (n – 1)d,
101 = 1 + (n – 1)(2)
∴ 101 = 1 + 2n – 2
∴ 101 = 2n – 1
∴ 102 = 2n
∴ n = `102/2`
∴ n = 51
Now, `S_n = n/2 (t_1 + t_n)`
∴ `S_51 = 51/2 (1 + 101)`
= `51/2 (102)`
= 51 × 51
= 2601
∴ The sum of odd natural numbers from 1 to 101 is 2601.
APPEARS IN
संबंधित प्रश्न
The first and the last terms of an AP are 8 and 65 respectively. If the sum of all its terms is 730, find its common difference.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, .... as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take `pi = 22/7`)
[Hint: Length of successive semicircles is l1, l2, l3, l4, ... with centres at A, B, A, B, ... respectively.]
Find the sum of the following arithmetic progressions:
3, 9/2, 6, 15/2, ... to 25 terms
Find the sum of first n odd natural numbers
Find the sum of all odd numbers between 100 and 200.
Find the sum of the first 15 terms of each of the following sequences having the nth term as
bn = 5 + 2n
Find the sum of the first 15 terms of the following sequences having nth term as yn = 9 – 5n.
The 7th term of an AP is –4 and its 13th term is –16. Find the AP.
If an denotes the nth term of the AP 2, 7, 12, 17, ..., find the value of (a30 - a20).
Which term of the AP 21, 18, 15, ... is zero?
The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.
If `4/5`, a, 2 are three consecutive terms of an A.P., then find the value of a.
Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is
Q.3
Q.16
Find whether 55 is a term of the A.P. 7, 10, 13,... or not. If yes, find which term is it.
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`.
Find S10 if a = 6 and d = 3.
The sum of all two digit odd numbers is ______.
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
