Q. Progression
10+84+734+.....= ?
Option
a)(9(9^n+1)/10)+1
b)(9(9^n-1)/8)+1
c)(9(9^n-1)/8)+n
d)(9(9^n-1)/8)+n^2

• 10+84+734+....
  =(9+1)+(9^2+3)+(9^3+5)+...
  =(9+9^2+9^3+...)+(1+3+5+7+...)
  =9(9^n-1)/(9-1)+n^2 [sum of GP=a(r^n -1)/(r-1) & sum of first n odd no.=n^2]
  =9(9^n-1)/8)+n^2
  option(d)
• 11 years agoHelpfull: Yes(12) No(1)
• ans is option (d) becoz put the value of n=1,2,3...
  n=1,(9(9^1-1)/8)+1^2 = 9(8/8)+1 = 10
  n=2,(9(9^2-1)/8)+2^2 = 9(80/8)+4 = 94 = 10+84
  n=3,(9(9^3-1)/8)+3^2 = 9(728/8)+9 = 9*91 + 9 = 819+9 = 828=10+84+734
  similarly..........
  
• 11 years agoHelpfull: Yes(3) No(0)
• ans is d
  n=1
  (9(9^1-1)/8)+1^2 = 9(8/8)+1 = 10
  
  n=2
  (9(9^2-1)/8)+2^2 = 9(80/8)+4 = 94 ( 10+84 )
  
  n=3
  (9(9^3-1)/8)+3^2 = 9(728/8)+9 = 9*91 + 9 = 819+9 = 828 (10+84+734)
  
  so option d is correct
• 11 years agoHelpfull: Yes(1) No(0)
• (9^1+9^2+9^3+...)+(1+3+5....)
  (9(9^n-1)/8)+(1+(n-1)2)
  (9(9^n-1)/8)+(2n-1)
• 11 years agoHelpfull: Yes(0) No(0)
• 10+84+734+....
  =(9+1)+(9^2+3)+(9^3+5)+...
  =(9+9^2+9^3+...)+(1+3+5+9+...)
  =9(9^n-1)/(9-1)+n^2 (sum of first n odd no.=n^2)
  =9(9^n-1)/8)+n^2
  option(d)
• 11 years agoHelpfull: Yes(0) No(0)
• d is the answer
• 11 years agoHelpfull: Yes(0) No(0)
• by putting value we can check
  n=1
  we get 10 in only one option (d) so(d)is right ans.
• 11 years agoHelpfull: Yes(0) No(0)
• apply the a.p. and g.p.
  ans -d
• 10 years agoHelpfull: Yes(0) No(0)