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Maths Puzzle
Numerical Ability
Which term of the series 5, 8, 11, 14,.....is 320?
A. 104th B. 105th
C. 106th D. 64th
Read Solution (Total 6)
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- its an A.P. series
an=a+(n-1)d where an(last term)=320;a(first term)=5;d(diff of two nos,)=3
320=5+(n-1)3
315/3= n-1
n=106
ans is c part - 10 years agoHelpfull: Yes(0) No(0)
- its ap series..
l=a+(n-1)d
l=last term
n=nth
d=diff
320=5+(n-1)3
n=106 - 10 years agoHelpfull: Yes(0) No(0)
- it is an A.P
Tn = a+(n-1)d
=> 320 = 5 + (n-1)3
=> 315 = (n-1)3
=> (n-1) = 105
=> n=106 - 10 years agoHelpfull: Yes(0) No(0)
- 320 = 5 + (n-1)3
315 = 3(n-1)
n = 106 - 10 years agoHelpfull: Yes(0) No(0)
- 320=5+(n-1)*3
n=106 - 10 years agoHelpfull: Yes(0) No(0)
- Ans : option C
This can be found using arithmetic progression. In this first term is a,(a+d), (a+2d), (a+3d) and so on it goes.
From the question we get,
a = 5 ---------first term,
(a+d) = (5 + 3 = 8) ----------second term,
(a + 2d) = ( 5 + (2*3)) = 11 ------------third term
therefore 106th term is ( a + 105d) = ( 5 + (105*3)) = 320
ans : 106th term is 320 - 10 years agoHelpfull: Yes(0) No(0)
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