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?

Options
1) 620
2) 625
3) 635
4) 615
5) 623

• General representation of an odd number is 2x+1
  so 9 consecutive odd numbers (ie set A) can be represented as
  2x+1 , 2x+3 , 2x+5, 2x+7, 2x+9 , 2x+11 ,2x+13 , 2x+15 ,2x+17
  It is told that sum of 9 consecutive odd numbers is 657
  so upon adding and equating
  (2x+1) + (2x+3) + (2x+5) +( 2x+7) + (2x+9) +( 2x+11) +(2x+13)+ ( 2x+15) +(2x+17)=657
  ie 18x+81 = 657
  18x =657-81
  x =576/18
  =32
  Now that we have found that x=32
  2x+1 is the 1st odd number. So upon substituting x we get
  first odd number in set A as 65
  We need to find the sum of 7 consecutive odd numbers whose lowest is 18 more than the lowest of set A.
  So the new starting odd number of the series will be 18+65=83
  so 83+85+87+89+91+93+95=623
  Therefore option 5
  
  
• 7 years agoHelpfull: Yes(11) No(0)
• n Set A, 9x + 2(1 + 2 + ..... + 8) = 657
  9x + 8*9*2/2 = = 657
  9x + 36 × 2 = 657
  x = 65
  
  lowest number of another set = 65 + 18 = 83
  
  Sum of seven consecutive number of another set
  = 7x + (1 + 2 + 3 + 4 + 5 + 6) × 2
  = 7x + 6*7*2/2
  = 7x + 21 = 83 × 7 + 42 = 623
• 8 years agoHelpfull: Yes(6) No(5)
• s=n/2[2a+(n-1)d] = 657 for set a
  s=657 , n=9 , d=2 (since odd number)
  find a , a=72.11
  Now, add 18 according to the problem.. to the loest number...vwhich becomes 90.11
  Again , S=7/2[2a+(n-1)d] = 3.5[2x90.11+(7-1)2] = 672.77
• 7 years agoHelpfull: Yes(2) No(5)