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The sum of five consecutive odd numbers of set p is 435. What is the sum of five consecutive numbers of another set q. Whose largest number is 45 more than the largest number of set p? |
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Answer» 670 Let the five consecutive ODD NUMBERS of set p be 2n - 3, 2n - 1, 2n + 1, 2n + 3, 2n + 5. Sum of these five numbers = 2n - 3 + 2n - 1 + 2n + 1 + 2n + 3 + 2n + 5 = 10N + 5 = 435 => n = 43 Largest number of set p = 2(43) + 5 = 91 The largest number of set q = 91 + 45 = 136 => The five numbers of set q are 132, 133, 134, 135, 136. Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670. |
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