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I416. The sum of the squares of two consecutive multiples of 7 is 1225. Findthe multiples.65 |
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Answer» Let the first multiple is 7*k and next multiple is 7*(k+1)therefore 49*k^2 + 49*(k^2 + 2*k + 1) = 122549*(2*k^2 + 2*k + 1) = 12252*k^2 + 2*k + 1 = 25k^2 + k -12 =0k = 3 or -4(rejected)k=3 k+1 =4therefore 21 and 28 |
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