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find the sum of all three digit numbers which are multiples of 9 |
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Answer» First three-digit number which is a multiple of 9 = 108 Second three-digit number which is a multiple of 9 = 117 Third three-digit number which is a multiple of 9 = 126 … nth three digit number which is a multiple of 9 = 999 So, the arithmetic sequence is 108, 117, 126, …, 999 Common difference = a = 117 − 108 = 9 First term of the arithmetic sequence = a + b = 108 ⇒ 9 + b = 108 ⇒ b = 108 − 9 = 99 Therefore, xn=9n+99, where n is a natural number. ⇒ 999 = 9n + 99 ⇒ 999 − 99 = 9n ⇒ 900 = 9n ⇒ n = 100 Therefore, there are 100 three-digit numbers, which are the multiples of 9. We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term. ∴ Sum of its first 100 terms = 1/2 x 100 x (x100+x1) … (1) Here, x1 = 108 and x100 = 999 Putting the values in equation (1): Sum of first 100 terms = 1/2 x 100 x (999+108) = 50 x 1107 = 55350 |
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