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In the sequence 1,2,2,3,3,3,4,4,4,4,....... Then what is 150th term

In the sequence 1,2,2,3,3,3,4,4,4,4,....... Then what is 150th term

Grade:11

4 Answers

Technical Hacks
29 Points
6 years ago
A sequence of numbers is given as: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 and so on. What will be the 100th term of that sequence?This question previously had question details. You can find them in the question comments.AnswerRequestFollow6Have this question too? Request Answers:Request From QuoraWe will distribute this question to writers, and notify you about new answers.Amar DoshiAmar Doshi, Maths Enthusiast98 Answers in MathematicsSandeep VijaySandeep Vijay, works at Cipla85 Answers in MathematicsBharat BhushanBharat Bhushan, former Executive Engineer at Larsen & Toubro Engineering (2014-2016)25 Answers in MathematicsANUJ KUMARAmitabha TripathiNeeraj DhimanView More or SearchPromoted by wishfin.comPersonal loans from 10.99%.Check your eligibility. Get offers from top 17 banks.Get Quote3 ANSWERSSahaj RamachandranSahaj Ramachandran, Math is my second nature.Answered Aug 2, 2015The answer is 14.The trick is to find a pattern. Consider numbering the given sequence as follows:1 is at position 12 is at positions 2, 33 is at positions 4, 5, 64 is at positions 7, 8, 9, 10and so on...Notice that the last positions of the individual values form a sequence.1,3,6,10,15,21...It`s the triangular series, where the Nth term is given by:n(n+1)2n(n+1)2, where n is the value.So for the value nn, it last appears at a position P=n(n+1)2P=n(n+1)2For example, 3 will last appear at 3∗(3+1)2=63∗(3+1)2=6Your question is to find the 100th term right? Then P=100P=100 and you have to find the value of nn for which P=100P=100 . Thus, P=n(n+1)2P=n(n+1)2 ⟹100=n(n+1)2⟹100=n(n+1)2Solving, we get the value as n=13.6509716980849n=13.6509716980849And the nearest integer is 14.Hope this helps.Thank You...Peace...
Roshan Jee Padhi
19 Points
6 years ago
1,2,2,3,3,3,3,4,4,4,4.......
1 is repeated once. 2 is repeated twice, 3 is repeated thrice and so on.
in  the 10th position is 4.
as you can clearly see it forms an AP. 1+2+3+4+10
in the same way, 1+2+3+.......n  terms=150
using the formulae Sn=n/2[2a+(n-1)d)
150=n/2[2+n-1]
=>300=n(n+1)
=>n2+n-300=0
Now it has taken the form of a quadratic equation
Solving it, using quadratic formulae, the roots = [-b+(b2-4ac)1/2]/2a
=(-1+​​12011/2)/2
the roots are -17.82 and 16.825. 
since the no. of terms i.e. n has to be positive, the ans is 16.825 or 17 (rounding it off to nearest integer).
therefore the 150th term in the original series would be 17.
 
 
nandini
17 Points
6 years ago
16 is ans
bcz sum of number of terms from 1 to 16 is 136 
136+17=153 
our question is 150th term then it must be 16...
 
Vansh
13 Points
5 years ago
n(n+1)/2>150
n²+n>300
Least value for n is 17
Hence 150th term is 17
as for putting n=17 the inequality becomes 306>300 and for n= 16 is not true

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