Advertisements
Advertisements
प्रश्न
Find the sum of all integers between 50 and 500, which are divisible by 7.
Advertisements
उत्तर
In this problem, we need to find the sum of all the multiples of 7 lying between 50 and 500.
So, we know that the first multiple of 7 after 50 is 56 and the last multiple of 7 before 500 is 497.
Also, all these terms will form an A.P. with the common difference of 7.
So here,
First term (a) = 56
Last term (l) = 497
Common difference (d) = 7
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
`a-n = a+ (n -1)d`
So, for the last term,
497 = 56 + (n -1)7
497 = 56+ 7n - 7
497 = 49 +7n
497 - 49 = 7n
Further simplifying
448 = 7n
`n = 448/7`
n = 64
Now, using the formula for the sum of n terms,
`S_n = n/2 [2a + (n -1)d]`
For n = 64 we get
`S_n = 64/2 [2(56) + (64 - 1) 7]`
= 32 [112 + (63)7]
= 32(1123 + 441)
= 32(112 + 441)
= 32(553)
= 17696
Therefore the sum of all the multiples of 7 lying betwenn 50 and 500 is `S_n = 17696`
APPEARS IN
संबंधित प्रश्न
The ratio of the sum use of n terms of two A.P.’s is (7n + 1) : (4n + 27). Find the ratio of their mth terms
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)]
If the pth term of an A. P. is `1/q` and qth term is `1/p`, prove that the sum of first pq terms of the A. P. is `((pq+1)/2)`.
Which term of the A.P. `20, 19 1/4, 18 1/2, 17 3/4,` ..... is the first negative term?
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.
If the numbers (2n – 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and times the corresponding sides of ΔPQR?
Write an A.P. whose first term is a and common difference is d in the following.
a = –3, d = 0
Choose the correct alternative answer for the following question .
In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....
If \[\frac{1}{x + 2}, \frac{1}{x + 3}, \frac{1}{x + 5}\] are in A.P. Then, x =
The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is
The common difference of the A.P.
Q.5
Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.
If the sum of first n terms of an AP is An + Bn² where A and B are constants. The common difference of AP will be ______.
The middle most term(s) of the AP: -11, -7, -3,.... 49 is ______.
Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.
[Hint (iii) : These numbers will be : multiples of 2 + multiples of 5 – multiples of 2 as well as of 5]
Find the sum of the integers between 100 and 200 that are not divisible by 9.
