Prove that arithmetic sequence 7,11,15etc.. does not contains perfect square
Answers
Step-by-step explanation:
a=7
d= 11-7=4
an = a+ (n-1)d
= 7 + 4n-4
= 4n +3
Let p be a natural number.
p^2 = 4n+3
n = p^2-3/4
Now... for all values of p from 1 to 9..
n does not come out to be an integer...
Hence.. The sequence doesn't have a perfect square....
Hope this helps....
Bhuvan..
Given : arithmetic sequence 7,11,15etc..
To find : prove that arithmetic sequence does not contains perfect square
Solution:
a = 7
d = 4
aₙ = a + (n - 1) d
= 7 + (n - 1) 4
= 7 + 4n - 4
= 4n + 3
Perfect square
k²
k = 2m or 2m + 1
k = 2m
k² = 4m² = 4n ≠ 4n+ 3
k = 2m + 1
k² = (2m + 1)² = 4m² + 4m + 1
= 4m(m + 1) + 1
= 4n + 1 ≠ 4n+ 3
Hence arithmetic sequence 7,11,15 does not contains any perfect square
Learn more:
If x^2 +39 is a perfect square then number of integral values of x is ...
https://brainly.in/question/17362511
A perfect square number has four digits none of which is zero. The ...
https://brainly.in/question/9447785