NCERT Class XI Mathematics - Sequences and Series - Solutions
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Question : 16
Total: 106
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Solution:
We have to find 105 + 110 + 115 + .......+ 995
This is an A.P. with first term a = 105, common difference d = 110 – 105 = 5 and last term l = 995
∴ l = a + (n – 1)d ⇒ 995 = 105 + (n – 1)5
⇒ 5n – 5 = 890 ⇒ 5n = 895 ⇒ n = 179
∴ Sum =
This is an A.P. with first term a = 105, common difference d = 110 – 105 = 5 and last term l = 995
∴ l = a + (n – 1)d ⇒ 995 = 105 + (n – 1)5
⇒ 5n – 5 = 890 ⇒ 5n = 895 ⇒ n = 179
∴ Sum =
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