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Numerical Ability
Sequence and Series
Find the 100th term of the following series - 3,6,12,21,33,48...
Read Solution (Total 13)
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- The given series i.e 3,6,12,21,33......
is derived from a(n+1)=a(n)+3n for n>0
where a(1)=3
now a(99+1)=a(100)=a(99)+3*99
a(98+1)=a(99)=a(98)+3*98
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a(1+1)=a(2)=a(1)+3*1
therefore
a(100)=a(99)+3*99
a(100)=a(98)+3*98+3*99
a(100)=a(97)+3*97+3*98+3*99
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continue in this way
finaly we get
a(100)=a(1)+3*1+3*2+3*3+.................+3*99
=3+ 3(1+2+3+4+5+6+7+8.....+99)
=3+3(sum of 99 A.P terms)
=3+3(99x100)/2
=14,853
- 12 years agoHelpfull: Yes(27) No(3)
- 14853.
1st term=3
2nd term=3+3*1
3rd term=3+3*1+3*2
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100th term=3+3*1+3*2.........+3*99
=3+3{1+2+3+.......99}
=3+3{(100*99)/2}
=14853
- 12 years agoHelpfull: Yes(26) No(1)
- 3=3*1;
6=3*(1+1);
12=3*(1+1+2);
21=3*(1+1+2+3);
33=3*(1+1+2+3+4);
48=3*1(1+1+2+3+4+5)& so on.. now the 100th term should b equal to=3*(1+the sum of (100-1)th natural no i.e.99*100/2=4950)=14853 - 12 years agoHelpfull: Yes(6) No(0)
- 3 6 12 21 33 48
take 3 common( 1 2 4 7 11 16.....)
now see the diff b/w no's (1 2 3 4 5.....)
now apply A.P formula tn=a+(n-1)d
t100= 1+(100-1)1
t100=100
now multiply that 3 which we took common so ans is 300 - 12 years agoHelpfull: Yes(5) No(19)
- @sakthi..u r finding the 100th term for series of common differences and not for the given series!!...we get 100th term in common difference as 100
- 12 years agoHelpfull: Yes(3) No(1)
- ans should be 14853
because t(n+1)=t(n)+3(n-1) - 12 years agoHelpfull: Yes(3) No(1)
- 14853 is the right answer
- 12 years agoHelpfull: Yes(3) No(0)
- ans is 15153
- 12 years agoHelpfull: Yes(2) No(6)
- its adiing like..
3,6,9..
so last term will be 48+18=66 - 12 years agoHelpfull: Yes(1) No(7)
- 3+3*0=3
3+3*0+3*1=6
3+3*0+3*1+3*2=12
3+3*0+3*1+3*2+3*3=21
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3+3(1+2+3+....+99)
3+3((99*(99+1)/2))
3+14850
14853...ans - 9 years agoHelpfull: Yes(1) No(0)
- vivek garu can u expn ur aans plzzzz????
- 12 years agoHelpfull: Yes(0) No(3)
- SAGAR , s i'm wrong, thanks for ur feedback
- 12 years agoHelpfull: Yes(0) No(0)
- ans is 66
- 12 years agoHelpfull: Yes(0) No(0)
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