options
a)24
b)25
c)26
d)27
d)27
Q: 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23 find the next term
options
a)24
b)25
options
a)24
b)25
@iLoveTorres said:Q: 17, 19, 21, 22, 23 find the next term
29??????
ese questions to bs randomly hojte h 
@sameersapre23 said:29??????ese questions to bs randomly hojte h
no.. chalo mein edit karke options daal deta hoon..
hint: ekdum bekaar logic hai.. soch bhi nahi paoge :P
hint: ekdum bekaar logic hai.. soch bhi nahi paoge :P
@Dexian said:29 kaise??
yr
17, 19 , 21, 22, 23
i am not sure of answer what logic i applied is that
1 st two are prime number in the series., and after that next two terms are product of prime numbers
21= 3*7
22= 2*11
and then again their should be prime numbers in next two terms
23 ,29
@iLoveTorres said:no.. chalo mein edit karke options daal deta hoon.. hint: ekdum bekaar logic hai.. soch bhi nahi paoge
u have edited the options?this is not fair
wese mera logic dekh neeche
f(x)= root(x+root(x-root(x+root(x............infinity)))) --------hope this is clear
find f(3) +f(7) +f(13)+f(21)
@iLoveTorres said:Q: 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23 find the next termoptions a)24b)25c)26d)27
25.
@amresh_maverick said:f(x)= root(x+root(x"-"root(x+rootx............infinity))) --------hope this is clearfind f(3) +f(7) +f(13)+f(21)
bhai yaha pe "-" sign hi hai? kyun ki then its not possible to solve
@iLoveTorres said:bhai yaha pe "-" sign hi hai? kyun ki then its not possible to solve
yes minus is there and then the series repeats itself
[] ----greatest Integer fun
[5/2] + [5/2+1/100] +[5/2+2/100] ................[5/2+199/100] =???
@amresh_maverick said:[] ----greatest Integer fun[5/2] + [5/2+1/100] +[5/2+2/100] ................[5/2+199/100] =???
600 ?..
2*50 + 3*100 + 4*50 = 600
@amresh_maverick said:f(x)= root(x+root(x-root(x+root(x............infinity)))) --------hope this is clearfind f(3) +f(7) +f(13)+f(21)
Let's solve for f(3) first..
Let f(3) = a = rt(3 + rt(3 - rt(3 + rt(3 - rt(3 + ...)))))
Let b = rt(3 - rt(3 + rt(3 - rt(3 + ...))))
So we can write a = rt(3 + b) i.e. a^2 = 3 + b
And b = rt(3 - a) i.e. b^2 = 3 - a
Subtracting the two equations we get
(a + b)(a - b - 1) = 0
i.e. b = a - 1 as (a + b) cannot be zero.
Till this point it is common solution for a being f(3) or f(7) or f(13) or f(21).
Putting the bold part in above quadratic to get an equation in a, we get
a^2 = 3 + a - 1
i.e. a^2 - a - 2 = 0
i.e. (a - 2)(a + 1) = 0
So a = f(3) = 2 as a is positive.
Now if a = f(7), slight change will be in above quadratic i.e. a^2 = 7 + a - 1
i.e. a^2 - a - 6 = 0
i.e. (a - 3)(a + 2) = 0
So a = f(7) = 3 as a is positive.
Similarly for a = f(13), we get a^2 = 13 + a - 1
i.e. a^2 - a - 12 = 0
i.e. (a - 4)(a + 3) = 0
So a = f(13) = 4 as a is positive.
And for a = f(21), we get a^2 = 21 + a - 1
i.e. a^2 - a - 20 = 0
i.e. (a - 5)(a + 4) = 0
So a = f(21) = 5 as a is positive.
Thus desired sum is 2 + 3 + 4 + 5 = 14 :)
Team BV
@amresh_maverick said:[] ----greatest Integer fun[5/2] + [5/2+1/100] +[5/2+2/100] ................[5/2+199/100] =???
[2.5] + [2.501] + [2.502] + ... + [2.99] + [3] + [3.01] + ... + [3.49] + [3.5] + ... + [4] + .. + [4.49]
2+2+2...(50 times) + 3+3+3+...(100 times) + 4+4+4... (50 times) = 600 ??
One from my side:
Find the number of distinct integers in the sequence: [1^2/2013], [2^2/2013], [3^2/2013], ..., [2013^2/2013].
@amresh_maverick said:[] ----greatest Integer fun[5/2] + [5/2+1/100] +[5/2+2/100] ................[5/2+199/100] =???
600.
(50* 2 + 100*3 + 50*4 )..
(50* 2 + 100*3 + 50*4 )..
@bodhi_vriksha said:One from my side: Find the number of distinct integers in the sequence: [1^2/2013], [2^2/2013], [3^2/2013], ..., [2013^2/2013].
[45^2/2013]------1
[64^2/2013]------2
..... [2013^2/2013] --------2013
distinct integers to saare aa jayegnge ??
[64^2/2013]------2
..... [2013^2/2013] --------2013
distinct integers to saare aa jayegnge ??