The 15th term in the given A.P. sequence is a₁₅ = 33.
According to the statement
we have given that the A.P. Series with the a = 5 and the d is 2.
And we have to find the 15th term of the sequence.
So, for this purpose we know that the
An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
And the formula is a
an = a + (n-1)d
After substitute the values in it the equation become
an = 5 + (15-1)2
a₁₅ = 5 + 28
Now the 15th term is a₁₅ = 33.
So, The 15th term in the given A.P. sequence is a₁₅ = 33.
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4n/4n-4 × n-1/n+1
= 4n(n-1)/(4n-4)(n+1)
= (4n^2 - 4n)/(4n^2 + 4n - 4n-4)
= (4n^2 - 4n)/(4n^2 - 4)
= 4(n^2 - n)/4(n^2 - 1)
=(n^2 - n)/(n^2 - 1)
The plus-minus sign represents that there are two possible outcomes.
In this case, we have

. When we branch out the possibilities we got 2 values:

and

Those are the roots of this equation. When they ask their product, they want you to multiply both numbers.
When we multiply them:

When we FOIL the we get:

Simplify:


So the product of the two roots of this equation is 6.
Setting AC = BD because the diagonals are congruent, then 3y/5 = 3y -4
y=5/3
So then AC = 1 = BD