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
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.
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