Advertisements
Advertisements
प्रश्न
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]
Advertisements
उत्तर
Since, multiples of 2 or 5 = Multiples of 2 + Multiples of 5 – [Multiples of LCM (2, 5) i.e., 10],
∴ Multiples of 2 or 5 from 1 to 500
= List of multiples of 2 from 1 to 500 + List of multiples of 5 from 1 to 500 – List of multiples of 10 from 1 to 500
= (2, 4, 6,..., 500) + (5, 10, 15,..., 500) – (10, 20,..., 500) ...(i)
All of these list form an AP.
Now, number of terms in first list,
500 = 2 + (n1 – 1)2
⇒ 498 = (n1 – 1)2 ...[∵ a1 = a + (n – 1)d]
⇒ n1 – 1 = 249
⇒ n1 = 250
Number of terms in second list,
500 = 5 + (n2 – 1)5
⇒ 495 = (n2 – 1)5 ...[∵ l = 500]
⇒ 99 = (n2 – 1)
⇒ n2 = 100
And number of terms in third list,
500 = 10 + (n3 – 1)10
⇒ 490 = (n3 – 1)10
⇒ n3 – 1 = 49
⇒ n3 = 50
From equation (i),
Sum of multiples of 2 or 5 from 1 to 500
= Sum of (2, 4, 6,..., 500) + Sum of (5, 10,..., 500) – Sum of (10, 20,..., 500)
= `n_1/2[2 + 500] + n_2/2 [5 + 500] - n_3/2[10 + 500]` ...`[∵ S_n = n/2(a + l)]`
= `(250/2 xx 502) + (100/2 xx 505) - (50/2 xx 510)`
= (250 × 251) + (505 × 50) – (25 × 510)
= 62750 + 25250 – 12750
= 88000 – 12750
= 75250
APPEARS IN
संबंधित प्रश्न
How many terms of the A.P. 65, 60, 55, .... be taken so that their sum is zero?
If the mth term of an A.P. is 1/n and the nth term is 1/m, show that the sum of mn terms is (mn + 1)
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Find the sum of the following arithmetic progressions:
1, 3, 5, 7, ... to 12 terms
Find the sum of all odd numbers between 100 and 200.
Find the sum of the first 40 positive integers divisible by 3.
Find the value of x for which the numbers (5x + 2), (4x – 1) and (x + 2) are in AP.
Write the next term of the AP `sqrt(2), sqrt(8), sqrt(18),`....
Find the first term and common difference for the A.P.
127, 135, 143, 151,...
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is \[\frac{\left( a + c \right) \left( b + c - 2a \right)}{2\left( b - a \right)}\].
Find the sum: 1 + 3 + 5 + 7 + ... + 199 .
The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P.
Which term of the sequence 114, 109, 104, ... is the first negative term?
Write the sum of first n odd natural numbers.
Q.4
Find the value of x, when in the A.P. given below 2 + 6 + 10 + ... + x = 1800.
In an A.P. sum of three consecutive terms is 27 and their products is 504. Find the terms. (Assume that three consecutive terms in an A.P. are a – d, a, a + d.)
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
The sum of A.P. 4, 7, 10, 13, ........ upto 20 terms is ______.
Which expression gives the sum of the first n even natural numbers \[2 + 4 + \cdots + 2n\]?
