Question

Find the Sum of those integers from 1 to 500 which are multiples of 2 or 5.

Answer 1

We know that,

Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2, 5)

Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (10)

Multiples of 2 or 5 from 1 to 500 = List of multiple of 2 from 1 to 500 + List of multiple

of 5 from 1 to 500 – List of multiple of 10 from 1 to 500

= (2, 4, 6… 500) + (5, 10, 15… 500) – (10, 20, 30… 500)

Required sum = sum(2, 4, 6,…, 500) + sum(5, 10, 15,…, 500) – sum(10, 20, 30,., 500)

Consider the first series,

2, 4, 6, …., 500

First term, a = 2

Common difference, d = 2

Let n be no of terms

a_n = a + (n – 1)d

500 = 2 + (n – 1)²

498 = (n – 1)²

n – 1 = 249

n = 250

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_1,

S₁ = S₂₅₀ = (250/2) ×[2+500]

S₁ = 125(502)

S₁ = 62750 … (1)

Consider the second series,

5, 10, 15, …., 500

First term, a = 5

Common difference, d = 5

Let n be no of terms

By nth term formula

a_n = a + (n – 1)d

500 = 5 + (n – 1)

495 = (n – 1)5

n – 1 = 99

n = 100

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_2,

S₂ = S₁₀₀ = (100/2) ×[5+500]

S₂ = 50(505)

S₂ = 25250 … (2)

Consider the third series,

10, 20, 30, …., 500

First term, a = 10

Common difference, d = 10

Let n be no of terms

a_n = a + (n – 1)d

500 = 10 + (n – 1)10

490 = (n – 1)10

n – 1 = 49

n = 50

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_3,

S₃ = S₅₀ = (50/2) × [2+510]

S₃ = 25(510)

S₃ = 12750 … (3)

Therefore, the required Sum, S = S₁ + S₂ – S₃

S = 62750 + 25250 – 12750

= 75250