2n = n+2
n = 1
p = 1 -----> not a prime number
This contradicts original statement.
So assumption is wrong.
Therefore, sum of first 'n' consecutive positive integers is not equal to the square of a prime number.
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Q3.
Pn = (3n²-n)/2 = n(3n-1)/2
When n is odd, 3n-1 is even
Therefore Pn can be rewritten as product of two integers:
Pn = n * (3n-1)/2 or
Pn = n/2 * (3n-1)
Now the only way that Pn can be prime, is if one of the factors is = 1
Case 1: n odd ----> Pn = n * (3n-1)/2
n = 1, (3n-1)/2 = 1, Pn = 1 -----> not prime
(3n-1)/2 = 1, n = 1, Pn = 1 -----> not prime
In all other cases (n ≥ 3), n > 1, (3n-1)/2 > 1, so Pn is not prime
Case 2: n even ----> Pn = n/2 * (3n-1)
n/2 = 1, n = 2, (3n-1) = 5, Pn = 5 -----> prime
(3n-1) = 1, n=2/3 (not an integer)
In all other cases (n ≥ 4), n/2 > 2, (3n-1) > 1, so Pn is not prime
The only value of n that gives Pn prime is n = 2 (P2 = 5)
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Q4
I'm not sure what you mean by "find the sizes of the triangle ABC"
What I can find, is the size of the angles of triangle ABC
AB = BC -----> Triangle ABC is isosceles -----> ∠BAC = ∠BCA = θ
AL bisects angle BAC -----> ∠LAC = θ/2
AL = AC -----> Triangle LAC is isosceles -----> ∠ALC = ∠ACL = ∠ACB = θ
∠LAC = 180 - 2θ
θ/2 = 180 - 2θ
5/2 θ = 180
θ = 2/5 * 180
θ = 72
In triangle ABC:
∠BAC = ∠BCA = 72
∠ABC = 180 - 72 - 72 = 36
In triangle LAC:
∠ALC = ∠ACL = 72
∠LAC = 180 - 72 - 72 = 36 = 1/2 ∠BAC