1,7,8,49,50,56,57,343,344,350,351,392,393,399,400……
The above sequence contains su m of distinct power of 7 in increasing order(7^0,7^1,7^1+7^0,7^2 etc.)What is value of term number 38 ?
16863,16864,17150,16857

• just represent the 38 in binary form...i.e. 100110
  now convert it into base 7 = 7^5+7^2+7
  just add them or check the unit digit(3)
  So the ans z 16863...:)
• 11 years agoHelpfull: Yes(44) No(1)
• 2^0th term=1st term=1=7^0
  2^1th term=2nd term=7=7^1
  3rd term=8=7^1+7^0
  2^2th term=4th term=49=7^2
  5th term=50=7^2+7^0
  6th term=56=7^2+7^1
  7th term=57=7^2+7^1+7^0
  according to this
  2^5th=32nd term is=7^5
  33rd is=7^5+7^0
  34th=7^5+7^1
  35th=7^5+7^1+7^0
  36th=7^5+7^2
  37th=7^5+7^2+7^0
  38th=7^5+7^2+7^1+7^0=16864(ans)
  
• 11 years agoHelpfull: Yes(8) No(6)
• 1,7,,8,49,50,56,57,343,344,350,351,392,393,399,400
  terms from starting where seven power (0,1,2,3....etc) are present is
  1st,2nd,4th,8th,16th,32th,64th..from starting...see the series here power of 7 is{0,1,2,3,4,5,6}respectively
  i.e term [1,2,4,8,16,32] = power to 7 {0,1,2,3,4,5]......eqn---(1)
  we need to find 38 th term
  38th term = 32 th term + 6th term from starting
  by eqn (1) 32 th term = 7^5 &&& 6th term from starting is 56
  so 38 th term = 7^5+56 = 16807+56
  16863 answer
• 11 years agoHelpfull: Yes(4) No(3)
• 38 term=19 term*7..again 19th term=(9th term*7)+1..
  In this way we get 16863...that is the 38th term..
• 11 years agoHelpfull: Yes(3) No(3)
• @riman: should'nt it be 7^6+7^3+7^2...
• 11 years agoHelpfull: Yes(1) No(1)
• 32th term =7^5.
  33th term =7^5+1
  34th term = 7^5+7
  35th term =7^5+8
  36th term =7^5+49
  37th trem=7^5+50
  38th term = 7^5+56=16863
• 5 years agoHelpfull: Yes(1) No(0)