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(vi)Find the sum of first 40 positive integers divisible by 6. |
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Answer» he first 40 positive integers divisible by 6 are 6,12,18,....... upto40 terms the given series is in arthimetic progression with first term a=6 and common difference d=6 sum of n terms of an A.p is n/2×{2a+(n-1)d} →required sum = 40/2×{2(6)+(39)6} =20{12+234} =20×246 =4920 |
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