IB DP Math MAA SL : IB Style Mock Exams – Set 10 Paper 2

Question

Consider the infinite geometric sequence 4480,  -3360, 2520, -1890, 000

(a) Find the common ratio, r.

(b) Find the 20th term.

(c) Find the exact sum of the infinite sequence.

Answer/Explanation

Ans:

(a) We have u1 = 4480   and u2 = -3360.

Hence the common ratio is 

\(r = \frac{u_{2}}{u_{1}}\)

\(=-\frac{3360}{4480}\)

= -0.75

(b) Using the nth term formula un = u1rn-1   with n = 20, we get

u20 = u1r20-1   

= (4480) (-0.75)20-1   

≈ -18.9

(c) Using the sum of an infinite geometric sequence formula \(S_{\infty }=\frac{u_{1}}{1-r},\) we find

\(S_{\infty }=\frac{4480}{1-(-0.75)}\)

=2560

Question

A bronze sphere has a radius of 10.5 cm.

(a) Find the volume of the sphere, expressing your answer in the form a × 10k,  1 ≤ a ≤ 10 and k ∈ Z+.

The sphere is to be melted down and remoulded into the shape of a cone with a height of 11.9 cm.

(b) Find the radius of the base of the cone, giving your answer correct to 3 significant figures.

Answer/Explanation

Ans:

(a) Using the volume of a sphere formula, we get

\(V_{sphere}=\frac{4}{3}\pi r^{3}\)

\(=\frac{4}{3}\pi (10.5)^{3}\)

≈ 4849.05 cm3

≈ 4.85 × 103 cm3

(b) Using the volume of a cone formula, we obtain

\(V_{cone}=\frac{1}{3}\pi r^{2}h\)

\(4849.5=\frac{1}{3}\pi r^{2}(11.9)\)                             [since Vcone   = V sphere]

r ≈  19.73 cm                        [by using G.D.C.]

r ≈  19.7 cm

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