Math, asked by TbiaSamishta, 1 year ago

the sum of nine consecutive odd number of set a is 657. what is the sum of seven consecutive odd numbers whose lowest number is 18 more than the lowest number of set a?

Answers

Answered by Sidyandex
2

The sum of 9 consecutive number of first set is 657.

Thus from the data we obtain the first set as { 65, 67, 69, 71, 73, 75, 77, 79, 81 }.

The lowest number of second set with 7 number is 18 more than the lowest of first set.

We have the second set as { 83, 85, 87, 89, 91, 93, 95 }.

The sum of the second set is 623.

Answered by TooFree
2

Define x:

Let x be the smallest odd number of the set.


The 9 consecutive odd numbers are:

x, (x + 2), (x + 4), ( x + 6), (x + 8), ( x + 10), (x + 12), (x + 14), (x + 16)


Solve x:

The sum of the set is 657

x + (x + 2) + (x + 4) + ( x + 6) + (x + 8) + ( x + 10) + (x + 12)+ (x + 14) + (x + 16) = 657

x + x + 2 + x + 4 +  x + 6 + x + 8 +  x + 10 + x + 12 + x + 14 + x + 16 = 657

9x + 72 = 657

9x = 585

x = 65


Find the smallest number in the set:

Smallest number = x = 65


Find the smallest number in the new set:

Smallest number is 18 more than the lowest lowest number

Smallest number = 65 + 18 = 83


Find the sum of 7 consecutive number whose lowest number is 83:

Sum = 83 + 85 + 87 + 89 + 91 + 93 + 95 = 623


Answer: The sum is 623


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