मराठी

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] - Mathematics

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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]

बेरीज
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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

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पाठ 5: Arithematic Progressions - Exercise 5.4 [पृष्ठ ५७]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
पाठ 5 Arithematic Progressions
Exercise 5.4 | Q 2.(iii) | पृष्ठ ५७

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