Advertisements
Advertisements
Question
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`.
Advertisements
Solution
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
∴ \[\boxed{136}\] = 4 + (n – 1) × 4
∴ 136 – 4 = (n – 1) × 4
∴ \[\boxed{132}\] = (n – 1) × 4
∴ `132/4` = n – 1
∴ 33 = n – 1
∴ n = \[\boxed{34}\]
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = `34/2 (4 + 136)`
∴ Sn = 17 × \[\boxed{140}\]
∴ Sn = \[\boxed{2380}\]
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is \[\boxed{2380}\].
APPEARS IN
RELATED QUESTIONS
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Find the sum of the following arithmetic progressions:
`(x - y)/(x + y),(3x - 2y)/(x + y), (5x - 3y)/(x + y)`, .....to n terms
Find the sum 25 + 28 + 31 + ….. + 100
If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.
Which term of the A.P. `20, 19 1/4, 18 1/2, 17 3/4,` ... is the first negative term?
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
How many numbers are there between 101 and 999, which are divisible by both 2 and 5?
Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.
The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.
HINT: Let these terms be (a – d), a, (a + d).
First term and the common differences of an A.P. are 6 and 3 respectively; find S27.
Solution: First term = a = 6, common difference = d = 3, S27 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]` - Formula
Sn = `27/2 [12 + (27 - 1)square]`
= `27/2 xx square`
= 27 × 45
S27 = `square`
In an A.P. the first term is – 5 and the last term is 45. If the sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
If m times the mth term of an A.P. is eqaul to n times nth term then show that the (m + n)th term of the A.P. is zero.
Write the sum of first n odd natural numbers.
If the sum of n terms of an A.P. is Sn = 3n2 + 5n. Write its common difference.
Mark the correct alternative in each of the following:
If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn − kSn−1 + Sn−2, then k =
The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
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
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find:
- the production in the first year.
- the production in the 10th year.
- the total production in 7 years.
If the first term of an A.P. is 5, the last term is 15 and the sum of first n terms is 30, then find the value of n.
