Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
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.
Find the sum of all even integers between 101 and 999.
Find the 8th term from the end of the AP 7, 10, 13, ……, 184.
The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term form the end is ( a + l ).
The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34. Find its common difference.
Choose the correct alternative answer for the following question .
What is the sum of the first 30 natural numbers ?
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.
The first term of an A. P. is 5 and the common difference is 4. Complete the following activity and find the sum of the first 12 terms of the A. P.
a = 5, d = 4, s12 = ?
`s_n = n/2 [ square ]`
`s_12 = 12/2 [10 +square]`
`= 6 × square `
` =square`
In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences.
If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term.
Write the common difference of an A.P. whose nth term is an = 3n + 7.
If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is
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
The common difference of an A.P., the sum of whose n terms is Sn, is
Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d. find A.P.
A sum of Rs. 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs. 20 less than its preceding prize; find the value of each of the prizes.
The sum of the first 15 multiples of 8 is ______.
If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)
Find the sum of last ten terms of the AP: 8, 10, 12,.., 126.
Measures of angles of a triangle are in A.P. The measure of smallest angle is five times of common difference. Find the measures of all angles of a triangle. (Assume the measures of angles as a, a + d, a + 2d)
