3 5 7 11 17
Therefore the next number in the list is 19. In the above calculation first 2 numbers added with 2 and last two numbers added with 3 but center number added with 4.
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Asked Feb 24 in Aptitude by AkshatMehta 410k points quantitative-aptitude.
. 7 4 11. Because 2 3 5 7 11 13 are the first 6 primes and then people could just put 17 19 23 without having to do any calculations. 5 2 7.
Therefore the correct option is b. Ways to form a committee - why is this approach incorrect. Find an answer to your question 2 3 5 7 11 x 17 nirajsharma5572 nirajsharma5572 16012020 Math Secondary School answered 2 3 5 7 11 x 17 1 See answer Advertisement Advertisement nirajsharma5572 is waiting for your help.
Rms 2 3 5 7 11 13 17 19 Have a question about using WolframAlpha. Im not sure why they didnt set T 6 13 its still solvable that way and it even gives you T 7 17. Explain why it is possible to conclude that 331 is prime without checking for mare prime factors Choose the correct answer below.
Sure dude whatever you say. From this they hide the number in this series 0213. The numbers 2 3 5 7 11 13 and 17 are not factors of 331.
Welcome to Sarthaks eConnect. 2 3 5 7 11 17 a 12 b 13 c 14 d 15. 17 The logic is.
The prime number that comes after 17 is 19. Clearly the given series are prime numbers starting from 2. 14 3 17.
11 3 14. The series given above is a prime number series - 3 5 7 1113 therefore the next number is 17. A prime number is a number which has only two factors-.
The prime number after 11 is 13. Correct Answer - Option 2. But 1 is neither a prime number nor a composite number.
Choose the correct alternative that will same pattern and fill in the blank spaces. Contact Pro Premium Expert Support. 2 is the only even prime number.
A prime number is a positive integer which has exactly 2 positive integer factore one factor being 1 and the other factor is the positive number itself. Choose the correct alternative. A set is given as S 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29.
So the next term in this sequence 2 3 5 7 11 13 17 would be 19. The series starts with 2 3 5 7 11 13 17 and 19 and goes on into infinity. The answer is 11.
N2n-1n2n-3cdotsn2n-6n-7 and because the sum of odd numbers between 1 and 2k1 is k12. 3 2 5. The dice is rolled 10 times and the results are added with given conditions.
This is a list of prime numbers. Now 33 and by induction we have proved that nn for all n geq 3. The series is 1 2 5 7 11 12 15 17.
A disaffected student might respond. It seems to be a list of non-composite odd natural numbers. What is the next number.
In following question a number series is given with one term missing. Answer 1 of 9. So the next prime numbers.
A unique platform where students can interact with teachersexpertsstudents to get solutions. Who cares Indeed primes were in the category. However the sequence 2 3 5 7 11 are the first five prime numbers.
4 9 5 12 7 15 8 x 10. A natural number can either be prime or composite. In following questiona number series is given with one term missing.
3 5 7 1113. Im not sure why they didnt set T 6 13 its still solvable that way and it even gives you T 7 17. 2 3 5 7 11 x 17.
As per this next number is 12 after 2 number should hide 13 14 and display 15 and then 1 number hide 16 then display 17. Because n2n is even we can write the odd numbers between the two polynomials to sum as. Hence the missing term is the prime number after 11.
So center number of 11 is not fit in the series. Add your answer and earn points. The square of 17 is 289 and the square of the next prime 19 is 361 so 331 is not divisible by any prime whose square is less than 331 O.
Since 11 and 22 the assertion is proved. What is the arithmetic mean of the set S.
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