Warm up your mind
Consider the sequence 1 , 2 , 2 , 3 , 3 , 3 , 4 , 4 , 4 , 4 , 5 , … where the integer n is repeated n times. What is the 1 5 0 th term?
The answer is 17.
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17 solutions
E x t r a : 0 ≤ n − ∫ 0 x X < X . . . . . . . . . . . . 0 ≤ 1 5 0 − ∫ 0 x X < X 0 ≤ 1 5 0 − ∫ 0 1 7 1 7 < 1 7 0 ≤ 3 < 1 7 1 7 − 3 = 1 4 t h e i n t e g e r ( 1 7 ) i n c u r r e n t t e r m r e p e a t e d ( 1 4 ) t i m e s , a n d t h e r e i s t h r e e m o r e ( 1 7 ) l e f t t i l l i t t u r n t o ( 1 8 ) .
it would be arithmetic or geometric progression
Pranit, can you explain your solution again, in a different way? It's a tad difficult for me to follow.
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I can certainly explain a more accessible method. Lets look at the positions of the last 1,2,3,... in the sequence. This comes out at 1,3,6,...,. From this we can see that this a sequence of triangle numbers. These can be calculated using an arithmetic sequence (as 1+2=3, 1+2+3=6 etc.), such that the nth term= n ( n + 1 ) / 2 = 150. We rearrange this to get the quadratic n 2 +n - 300=0, which can be solved to get 16.8...>16, so the answer is 17
Sum of 1 to 16 = 136 .
150 - 136 = 14
17 = 137
17 = 138
17= 139
And so on.. we get 17=150
Therefore 17 is the answer.
how to calculate sum of 1 to 16 in shortcut way ?
@Dollesin Joseph ...nice method but can you give a generalized way??
Here's my brief but nice solution.. By using Arithmetic Summation , S = 2 n [ 2 a 1 + d ( n − 1 ) ] Take a look again from the sequence, for every END of the integer n with repeated n times, there is a corresponding summation S , therefore.. 1 5 0 = 2 n [ 2 ( 1 ) + 1 ( n − 1 ) ] the value of n here is ranging somewhat 1 6 < n < 1 7 .. therefore final answer is 1 7
There is an even easier way than what others have posted. The N th term of the series is f ( N ) = ⌈ 2 x + 4 1 − 2 1 ⌉ where ⌈ ⌉ denotes the ceiling function.
Plugging in N = 1 5 0 gives 17.
how did you got this idea
Can you prove it?
adding n integer one time i.e 1+2+3+4....17=153
changing how we look at the sequence as follows: consider a serie made of the ranks of the original seq. we get s1=1 s2=3 s3=6 s4=10 and so on this is triangular serie with general expression n{n+1}/2 this should yield 150 so n{n+1}=300 and so n=17.82 or the 150th term is 17
1+2+3+4+5+6+........................................+16+17=152(as each no like 2 no has 2 no) But the last 17th terms will be 17 so the 150th term is also 17
If each number is n^nth where n is the value in the sequence, then all you do is 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17 til you reach over 150.Then take the current number and you have it.
The answer will be the minimum value of 'n' for which 'n(n+1)/2' is greater than 150.
first we add all number from 1 to 16. let z be the sum
z=
2
1
6
2
−
1
2
+
1
+
1
6
z=
2
2
5
6
+
1
6
z= 136
after that more 17 17s which goes to 153.
so, since 153>150>136 so 150th term is 17,
These are triangle numbers. So solving for n in 2 n ( n + 1 ) = 1 5 0 gives you ~16.82, which means we must be on "17, 17, 17, 17 ..."
n(n+1)/2=17(18/2)=153 153+2=155
by inspection we find that last "n" is n(n+1)/2 th term. so 17*18/2=153 that is last 17 would be the 153rd term of the sequence. and 150th would also be 17.
It would be better if he didnt write the pattern.. I just solved it for 40 seconds.
the number is repeting ntimes of n number so when you sum of 1 to 17 the answer is 153 & we asked the 150 th term so that is 17
Looking at the series we have a sum of all natural numbers, so n(n+1)/2=150 n^2+n-150=150 n=(1±√(1+1200))/2=17,8277 and -16,8277 So the answer =17 or -17 if you go the other (negative way )
Suppose a new number is introduced at (n+1)th term. Then n= summation of first n natural numbers. Therefore 150 is greater than or equal to sum of first m natural numbers. Therefore 150 is greater than equal n(n+1)/2. Therefore n(n+1) is less than 300. Therefore n=16. So the 150th term will be 17.