This is an arithmetic progression based question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise – 9.6
In this question we have to Find the sum of all integers from 1 to 500 which are multiples of 2 or 5.
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Understanding CBSE Mathematics
Class :- 10th
Question no 55(v)
Integers from 1 to 500 which are multiples of 2 are 2, 4, 6, 8, . . . .500.
First term(a) = 2, common difference(d) = 4 – 2 = 2 and nth term(an) = 500.
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
500 = 2 + (n – 1)2
=> 2(n–1) = 498
=> n–1 = 249
=> n = 250
Let S1 be the sum of this A.P. Hence, S1 = 250[2 + 500] / 2 = 125[502] = 62750.
Integers from 1 to 500 which are multiples of 5 are 5, 10, 15, 20, . . . .500.
First term(a) = 5, common difference(d) = 10 – 5 = 5 and nth term(an) = 500.
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
500 = 5 + (n – 1)5
=> (n – 1)5 = 495
=> n – 1 = 99
=> n = 100
Let S2 be the sum of this A.P. Hence, S2 = 100[5 + 500] / 2 = 50[505] = 25250.
Integers from 1 to 500 which are multiples of 2 as well as 5 are 10, 20, 30 . . . .500.
We know, 500 = 10 + (n – 1)10
=> 10(n – 1) = 490
=> n – 1 = 49
=> n = 50
Let S3 be the sum of this A.P. Hence, S3 = 50[10 + 500] / 2 = 25[510] = 12750.
Hence, required sum = S1 + S2 – S3
= 62750 + 25250 – 12750
= 75250