All three series converge, so the answer is D.
The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.
Consider a geometric sequence with the first term a and common ratio |r| < 1. Then the n-th partial sum (the sum of the first n terms) of the sequence is
Multiply both sides by r :
Subtract the latter sum from the first, which eliminates all but the first and last terms:
Solve for :
Then as gets arbitrarily large, the term will converge to 0, leaving us with
So the given series converge to
(I) -243/(1 + 1/9) = -2187/10
(II) -1.1/(1 + 1/10) = -1
(III) 27/(1 + 1/3) = 18
RTP: [a tan(u) + b]² + [b tan(u) - a]² = (a² + b²) sec²(u)
Proving LHS = RHS:
LHS = [a tan(u) + b]² + [b tan(u) - a]²
= a² tan²(u) + 2ab tan(u) + b² + b² tan²(u) - 2ab tan(u) + a²
= (a² + b²) tan²(u) + (a² + b²)
= (a² + b²)[tan²(u) + 1]
= (a² + b²) sec²(u), using the identity: tan²(x) + 1 = sec²(x)
= RHS
The answer is c, since the square root of 81 is positive or negative 9.
Hope this helps
150 * 1.25 * 1.25* 1.25 = £292.97
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