Question

Find the sum of all natural numbers lying between 100 and 1000, which ate multiples of 5 ​

Answer 1

Step-by-step explanation:

a¹ (First term) = 100 + 5

= 105

aⁿ ( last term ) = 1000 - 5

= 995

d ( Common

difference) = 5

n ( N0: of terms ) = [(aⁿ - a¹) / d] + 1

= (995 -105) /5 ] + 1

= (890/5) + 1

= 178 + 1

= 179//

∴ NO: OF TERMS (n) = 179

Xⁿ -------> Sum of all terms

Xⁿ =X / 2 [ a¹ + aⁿ ]

= 179/2 [ 105 + 995 ]

= 179/2 ( 1100)

= 179 × 550

= 98450 //

∴ SUM OF ALL NATURAL NUMBERS ( Xⁿ ) = 98450

Answer 2

✰ Cᴏʀʀᴇᴄᴛ Qᴜᴇsᴛɪᴏɴ :-

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5

✰ Given :-

• Numbers lying between 100 and 1000, which are multiples of 5.

✰ Tᴏ Fɪɴᴅ :-

• The sum of all natural numbers.

✰ Sᴏʟᴜᴛɪᴏɴ :-

The natural numbers lying between 100 & 1000, which are multiples of 5, are 105, 110, 115.....995

• a (1st term) = 105
• d (common difference) = 5

Now,

an = a + (n - 1) d

➙ 995 = 105 + (n - 1) × 5

➙ 890 = (n - 1) × 5

➙ 178 = (n - 1)

➙ n = 179

Sn = n/2 [2a + (n - 1) d]

➙ S179 = 179 = 179/2 [2 × 105 + (179 - 1) × 5]

➙ S179 = (179)[550]

➙ 98450

.°. The sum of all natural numbers lying between 100 & 100, which are multiples of 5 is 98450