First 3 terms are a^2 + n a^(n-)1 b + n(n-1)/2 * a^(n-2) b^2
So q^2 / pr = (n^2 * a^(2n-2) * b^2 ) / (1/2 * a^n * (n(n-1) * a^(n-2) * b^2 )
= n^2 * a^2n-2 * b^2
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1/2 n(n-1) * a^(2n-2) * b^2
= 2n / n - 1 as required
given p = 4, q=32 and r = 96:-
32^2 / 4*96 = 2n / n-1
2n / n-1 = 8/3
6n = 8n - 8
2n = 8
n = 4 answer
Answer:
x intercepts(-3/2,0)
y intercepts(0,-1)
Step-by-step explanation:
Answer:
3 and 4
Step-by-step explanation:
Consider squares on either side of 15, that is 9 and 16, so
< < , that is
3 < < 4
Answer:
Step-by-step explanation: The answer of this question is functional because of the dilemma of fractions make it functional.
-8/4=-2
x= -8
hope this helps!