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
The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of numbers of houses preceding the house numbered x is equal to sum of the numbers of houses following x.
How many terms of the A.P. 18, 16, 14, .... be taken so that their sum is zero?
In an AP Given a12 = 37, d = 3, find a and S12.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Find the sum of the following arithmetic progressions: 50, 46, 42, ... to 10 terms
Find the sum of the first 51 terms of the A.P: whose second term is 2 and the fourth term is 8.
Find the middle term of the AP 10, 7, 4, ……., (-62).
How many three-digit numbers are divisible by 9?
Write the next term for the AP` sqrt( 8), sqrt(18), sqrt(32),.........`
Find the sum (−5) + (−8)+ (−11) + ... + (−230) .
Write the nth term of an A.P. the sum of whose n terms is Sn.
If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to
If \[\frac{5 + 9 + 13 + . . . \text{ to n terms} }{7 + 9 + 11 + . . . \text{ to (n + 1) terms}} = \frac{17}{16},\] then n =
The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20,......
Activity :- Here, a = 12, d = `square`, n = 100, S100 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]`
S100 = `square/2 [24 + (100 - 1)"d"]`
= `50(24 + square)`
= `square`
= `square`
In a ‘Mahila Bachat Gat’, Sharvari invested ₹ 2 on first day, ₹ 4 on second day and ₹ 6 on third day. If she saves like this, then what would be her total savings in the month of February 2010?
If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is ______.
Rohan repays his total loan of ₹ 1,18,000 by paying every month starting with the first installment of ₹ 1,000. If he increases the installment by ₹ 100 every month, what amount will be paid by him in the 30th installment? What amount of loan has he paid after 30th installment?
