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Find the 100th term of the following series - 2,4,8,14,22,32...
Read Solution (Total 11)
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- t(n+1)=n^2+n+2
t(0+1)=0+0+2=2
t(1+1)=1^2+1+2
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t(99+1)=99^2+99+2=9902 - 12 years agoHelpfull: Yes(38) No(2)
- 2nd term -1st term=1*2
3rd term -2nd term=2*2
4th term -3rd term=3*2
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nth term -(n-1)th term=(n-1)*2
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sum:nth term -1st term=[1+2+3....(n-1)]*2(as rests are cancelld)
so 100th term=(100*99/2)*2+2(as sum=n(n-1)/2,n=100,1st term =2)
=9902ans - 12 years agoHelpfull: Yes(5) No(1)
- ANS IS : 9902
a(n) = 2 + n(n - 1)
a(100)=2+100(100-1)
=9902 - 9 years agoHelpfull: Yes(3) No(0)
- t(n)=t(n-1)+2n ;n>0
t(0)=2
t(1)-t(0)=2
t(2)-t(1)=4
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t(100)-t(99)=200
summing up we get t(100)=(2*100*101/2)+t(0)=10102 - 12 years agoHelpfull: Yes(2) No(7)
- 2 4 8 14 22 32
diff 2 4 6 8 10
again diff 2 2 2 2
now apply formula tn=a+(n-1)d
t100=2+(99)2
t100=200 - 12 years agoHelpfull: Yes(2) No(6)
- @sumit gupta----
actually the t(n) comes from the given logi....its not aaplicable for all questions..
s(n)=3+6+12+...+t(n)------eq 1
s(n)= 3+6+12+...+t(n)+t(n+1)------eq2
eq1-eq2
t(n+1)=3+(3*n)/2+(3*n*n)/2
this is d correct way....enjoy - 12 years agoHelpfull: Yes(1) No(1)
- @rupesh
Find the 100th term of the following series - 3,6,12,21,33,48...
now what will be the logic - 12 years agoHelpfull: Yes(0) No(0)
- can we solve the series 3,6,12,21,33.... using the above method i.e.
t(n+1) = n^2+ n+2? - 12 years agoHelpfull: Yes(0) No(1)
- @sumit...u rfinding the nth term for series of common differences and not for the given series!!
- 12 years agoHelpfull: Yes(0) No(0)
- .Now, How can you find any term like say 22 if you're given 2 only and this difference. Just the some of the differences plus 2.=> 2+ (2+4+6+8) = 22. Hence, 100th term = Sum of AP (2,4,6,8...) till 99 terms + 2=> [ 99*(2+98*2)/2 ] + 2 = 99*99 + 2 = 9801 + 2 = 9803.
- 12 years agoHelpfull: Yes(0) No(7)
- here nth term =2*n-1+n-1th term
so 100 term=2*99+99th term
99th term=2*98+98th term and so on
s0
2(99+98.....1)+2=9704 - 12 years agoHelpfull: Yes(0) No(3)
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