What is the next number of the following sequence 1.414, 1.732, 2.236, 2.646, 3.317, .....

1st number is=1.414=sqrt(2)
2nd number is=1.732=sqrt(3)
3rd number is=2.236=sqrt(5)
4th number is=2.646=sqrt(7)
5th number is=3.317=sqrt(11)
All sqrt numbers are prime number starting from 2
So, next number will be=sqrt(13)=3.60555
So, option 4)3.606 is the right answer

Here the series is the root of All prime numbers
starting from 2 onwards.
So 2,3,5,7,11,13....
So the next number will be sqrt(13)
which is 3.606
So option 4
That's it.

sqrt(2)=1.414
sqrt(3)=1.732
sqrt(5)=2.236
sqrt(7)=2.646
sqrt(11)=3.317
2,3,5,7,11 this is sequence of prime number
So the next term is 13. Therefore,
sqrt(13)=3.606

The given series is the roots of prime numbers
2^(1/2),3^(1/2),5^(1/2),7^(1/2) ............
so the answer is 13^(1/2)=3.606

series of square roots of prime numbers.
2,3,5,7,11,13
so answer is sqaureroo(13)=3.606

square root of consecutive prime numbers...

sq root of prime numbers

so, sqrt(13) = 3.606

Square root of prime numbers i.e.
Sqrt(2) , sqrt(3) , sqrt(5) , ...

=> Next term = sqrt(13) = 3.606

sqrt of 2, 3, 5, 7, 11, ....
next sqrt of 13=3.606
option 4

root 13

Concept is to write square root of numbers without producing rounded values....

square root (2) = 1.414
square root (3) = 1.732
skip square root (4) as it gives rounded value("2")
square root (5) = 2.236
.
.
.
square root (13) =3.606

That would be answer...

The given series is a square root of consecutive prime numbers
2,3,5,7,11,13..
Square root of 2=1.414
Square root of 3=1.732
Square root of 5=2.236
Square root of 7=2.646
Square root of 11=3.317
Square root of 13=3.606

so,answer is 3.606

2^1/2=1.414
3^1/2=1.732
5^1/2=2.236
7^1/2=2.646
11^1/2=3.317
13^1/2=3.606
Note: 2,3,5,7,11,13 are prime numbers
Answer: 4)3.606

Given sequence is 1.414, 1.732, 2.236, 2.646, 3.317,......
we have prime numbers 2,3,5,7,11,13,....
sqrt(2) = 1.414, sqrt(3) = 1.732, sqrt(5) = 2.236, sqrt(7) = 2.646, sqrt(11) = 3.317, and sqrt(13) = 3.606, ......
ans: 3.606

Series consists of Square root of prime numbers like 2,3,5,7,11.
next prime no. is 13.
so next number = Sqrt (13)= 3.606

it follows the order sq of 2,sq of 3 sq of 5 sq of 7 sq of 11 and the next will be sq of 13

sqrt(13)

root of all prime numbers are here in the series
2^.5=1.414
3^.5=1.732
5^.5=2.236
7^.5=2.646
11^.5=3.317
the next prime number is 13
13^.5=3.606

root 13= 3.60555
means approx 3.606

The series is of square root of 2,3,5,7,11. So next number will be root 13. As all the numbers are prime number in a sequence. So option 4 is correct.

square roots of the successive prime numbers starting from 2

Square root of 2 is = 1.3414
Square root of 3 is = 1.732
Square root of 5 is = 2.236
Square root of 7 is = 2.646
Square root of 11 is = 3.317
Square root of next prime number ,that is 13, is= 3.606

Its root of 13 simple

sqrt(2) = 1.414
sqrt(3) = 1.732
sqrt(5) = 2.236
sqrt(7) = 2.646
sqrt(11) = 3.317

If we look at series in the brackets, the numbers in bracket are prime numbers.

So the next number in the bracket will be 13 and answer will be

sqrt(13) = 3.606

The given below values are in series of square root of prime numbers i.e., sqrt(2),sqrt(3),sqrt(5),sqrt(7), sqrt(11) =3.137 then sqrt(13) = 3.606

Because it showing square root of prime numbers

square root of all prime nos

series is sqrt(prime number). so, next number is sqrt(13)=3.606, so option (4) is correct

sqrt(2),sqrt(3),sqrt(5),sqrt(7),sqrt(11),sqrt(13)
because it is series of prime number.

optio(4).

square root 13
series of square roots of prime no's

this is series of square root of prime numbers

it's series of squre root of prime number.

square root of consecutive primes...
so ans is sqrt of 13 which is 3.606

(2)^(1/2)=1.41.4
(3)^(1/2)=1.732
like this we get the squares of prime no.
so, (13)^(1/2)=3.606

ans 3.606

3.606....bcz given series are square root of prime numbers....like 2,3,5,7,11...
so next number of this series will be 13.

square root of 10

sq root of prime numbers.
so after 11, 13 is the next prime no.
sq root(13)=3.606 is the answer

values of the square roots of prime numbers.
sqrt(13)=3.606

given-sqrroot2,sqrroot3,sqrroot5,sqrroot7,sqrroot11
the series is of sqrroot of prime numbers and the next prime number is 13 and sqrroot13=3.606

Square root of primary numbers 2,3,5,7,11,13

ans= root 13
root2,root 3,root5,root7,root11,root13
2,3,5,7,11,13 are prime numbers

sqrt(13)

these are the square roots of prime no.
2,3,5,7,11,13....
sqrt(13)=3.606

The terms are under-roots of 2,3,5,7,11
So the next term must be 13^0.5

sqrt(2) = 1.414
sqrt(3) = 1.732
sqrt(5) = 2.236
sqrt(7) = 2.646
sqrt(11) = 3.317
sqrt(13) = 3.606
this series is of squareroot of prime numbers, after 11 , 13 is the next prime number, so next in the series is sqrt(13) i.e. 3.606

root over of prime numbers...
root over of 2,3,5,7,11,13
so option 4-3.606

root(13)

Square roots of odd numbers.
2,3,5,7,11,13
Therfore,sqrt(13)=3.6055

sqrt(13)

1.414 is square root of 2,
1.732 is square root of 3,
2.236 is square root of 5,
2.646 is square root of 7,
3.317 is square root of 11,
3.606 is square root of 13

Sq2, sq3, sq5, sq7, sq11, sq13 all prime number

SQUARE ROOTS SEQUENCE OF PRIME NO 'S
SQRT(2),SQRT(3).......SQRT(13)
SQRT(13)=3.606

option 4: 3.606
all the numbers are square root of prime numbers
sqrt(2)=1.414
sqrt(3)=1.732...... sqrt(13)=3.606

sqrt(13)=3.606

root 3= 3.6055

square root of number
1=1.000
2=1.414
3=1.732(prime no)
4=2.000
5=2.236(prime no)
6=2.449
7=2.646(prime no)
8=2.828
9=3.000
10=3.162
11=3.317(prime no)
12=3.404
answer 13=3.606(prime no)

sqrt(2)=1.414
sqrt(3)=1.732
sqrt(5)=2.236
sqrt(7)=2.646
sqrt(11)=3.317
sqrt(13)=3.606

sqrt(of prime)
sqrt(13)=3.606

the above answer is for prime nos root
root 2
root 3
root 5
root 7
root 11
and finally
root 13

The given series is The Squareroot of all the prime numbers.
sqrt(2)=1.414
sqrt(3)=1.732
sqrt(5)=2.236
sqrt(7)=2.646
sqrt(11)=3.317
sqrt(13)=3.606

square root of prime numbers 2,3,5,7,11,13

series of sqrt of prime nos. sqrt(13) = 3.606 (approx.) Ans.

all terms are the square root of consecutive prime no.
next term=root(13) =3.6055

all terms are the square root of consecutive prime no.
next term=root(13) =3.6055

the order is 1 3 5 7 11 13(primenumbers)...the above given numbers are the the roots of these ..so root of 13 is 3.606

simple sqaure root of prime number
so in this series the next number will be=(13)^0.5=3.606

all the given numbers in the sequence are the square root values of prime numbers i.e
2,3,5,7,11 thus nxt is root of 13=3.606

root 2 = 1.414 (rounded to three digits)
root 3 = 1.732
root 5 = 2.236
root 7 = 2.646
root 11 = 3.317

Hence we have to find the root of the next prime number as all the numbers are prime numbers

Nest prime is 13 and
root 13 is 3.606

squareroots of prime numbers

next no. will be sqrt(11)

sqrt 11

this is in the form sqrt2 , sqrt3, sqrt5, sqrt7, sqrt10

square root of prime numbers

i.e

sq rt of 2=1.414
sq rt of 3=1.732
sq rt of 5=2.236
sq rt of 7=2.646
sq rt of 11=3.317
sq rt of 13=3.606

Are there compilers out there that will intelligently optimize code for multiple processors? You might say that code HAS to be written for multiple processors, but a lot of code isn't that could be.
Consider a simple little app like this, written in crappy pseudocode. I put line numbers so you can see the steps involved.
1. Initialize a variable "a"
2. Initialize a variable "b"
3. Initialize a variable "c"
4. Initialize a variable "d"
5. Store 10 into a
6. Store 15 into b
7. Store 20 into c
8. Store 25 into d
9. Add a and b and store it to C
10. Multiply D with C and store it to A
Now, what if a compiler optimized this code for multiple processors? You see, the problem with multithreading is dependency on previous operations. What if a code skimmed through and realized that, for example, variable A wasn't being used in lines 2,3, 4, 6, 7, or 8? An optimized code for two processors could look something like this:
1. Initialze "a" and "b"
2. Initialize "c" and "d"
3. Store 10 into "a" and 15 into "b"
4. Add "a" and "b" and store it into "c"
5. Multiply "d" with "c" and store it into "a"
The whole idea is based on data dependency on previous and future operations.
Let's take a more extreme example:
1. Initialize "a"
2. Initialize "b"
3. Initialize "c"
4. Initialize "d"
5. Initialize "e"
6. Initialize "f"
7. Initialize "g"
8. Initialize "h"
9. Store 1.414 into "a"
10. Store 1.732 into "b"
11. Store 2.236 into "c"
12. Store 2.646 into "d"
13. Store 3.317 into "e"
14. Store 3.606 into "f"
15. Store 4.123 into "g"
16. Store 4.359 into "h"
17. Add "e" and "f" and store it to "g"
18. Add "a" and "b" and store it to "c"
19. Multiply "d" and "c" and store it to "a"
20. Multiply "h" and "g" and store it to "e"
This can be rewritten for two threads:
1. Initialize "a" ; Initialize "b"
2. Initialize "c" ; Initialize "d"
3. Initialize "e" ; Initialize "f"
4. Initialize "g" ; Initialize "h"
5. Store 1.414 into "a" ; Store 1.732 into "b"
6. Store 2.236 into "c" ; Store 2.646 into "d"
7. Store 3.317 into "e" ; Store 3.606 into "f"
8. Store 4.123 into "g" ; Store 4.359 into "h"
9. Add "e" and "f" and store it to "g" ; Add "a" and "b" and store it to "c"
10. Multiply "d" and "c" and store it to "a" ; Multiply "h" and "g" and store it to "e"
What do you think?