1.11 Infinite geometric series questions for IB Maths AI

Write down the common ratio, $r$.

Explain why the sum to infinity of the series exists.

Find the sum to infinity of the series.

Find the sum of the first 8 terms of the series.

Find $u_2$ and $u_3$, giving your answers in the form $a+b\mathrm{i}$.

Find the exact value of $\mathrm{\mid }r\mathrm{\mid }$.

(i) Find $\arg r$.

(ii) Hence find the smallest value of $n>2$ for which the pin $u_n$ lies in the first quadrant.

Show that the lengths of the pieces of thread, $\mathrm{\mid }{u}_2-u_1\mathrm{\mid }$, $\mathrm{\mid }{u}_3-u_2\mathrm{\mid }$, $\mathrm{\mid }{u}_4-u_3\mathrm{\mid }$, … form a geometric sequence with common ratio $\mathrm{\mid }r\mathrm{\mid }$.

Find the length of the first piece of thread, $\mathrm{\mid }{u}_2-u_1\mathrm{\mid }$.

Ana uses only pieces of thread that are at least $0.5$ cm long. Find the number of pieces of thread she uses.

Find the total length of thread that Ana uses.

Ana claims that, however many pins she used, she would never need more than $25$ cm of thread. Determine whether her claim is correct.

Find the monthly payment for Option A.

Find the total amount of interest Ines pays with Option A.

Find the amount Ines would still owe after 2 years with Option A.

Option B is advertised as a "flat-rate loan at only 4.2 % per year". For each year of the loan, interest equal to 4.2 % of the original amount borrowed is added to the debt, so the total debt after $n$ years, $u_n$, forms an arithmetic sequence. The total debt after 5 years is then repaid in 60 equal monthly payments.

Show that the total debt after 5 years is EUR 29 040.

Hence write down the monthly payment for Option B.

Find the nominal annual interest rate, compounded monthly, that Ines is actually being charged with Option B.

Ines says that Option B is cheaper than Option A because 4.2 % is less than 6.0 %. Explain why she is wrong.

Option C is a "growing payment" loan with the same interest rate as Option A, 6.0 % compounded monthly, which is 0.5 % per month. The first payment is EUR $P$, and each later payment is 1 % larger than the previous one, so that the payments form a geometric sequence.

Write down an expression, in terms of $P$ and $k$, for the $k$th payment.

The present value of a payment of $X$ made at the end of month $k$ is $\frac{X}{{1.005}^k}$.

Show that the sum of the present values of all 60 payments is

The loan is fully repaid when the sum of the present values of the payments equals the amount borrowed.

Hence find the value of $P$.

Find

(i) the size of the final payment;

(ii) the total amount Ines repays with Option C.

Option C costs more in total than Option A. Suggest one reason why Ines might still choose Option C.

Ines later estimates the value of the bakery as the sum of the present values of all its future monthly profits, again discounting at 0.5 % per month. She models the profit at the end of the first month as EUR 1500, with each month's profit 0.3 % greater than the previous month's, and assumes this continues indefinitely.

Find the value of the bakery according to this model.

Explain why this method of valuation could not be used if Ines assumed that the monthly profit grew by 0.6 % each month.

Find the common ratio of the sequence.

The advert will be withdrawn after the first week in which it reaches fewer than 500 new viewers. Find the number of weeks for which the advert runs.

Show that the present value of the first payment, made after one year, is 38 095.24 NOK.

The present values of the 1st, 2nd, 3rd, … payments form a geometric sequence.

(i) Find the present value of the 2nd payment.

(ii) Write down the common ratio of the sequence.

Find the total present value of the first 20 payments.

The council wants the payments to continue for ever. Find the amount that must be invested now.

Explain, in context, why the answer to part (d) is 20 times the annual payment.

To allow for rising costs, the council decides instead that each payment will be 2% larger than the previous one, starting with 40 000 NOK at the end of the first year. Find the amount that must be invested now for these payments to continue for ever.

State, with a reason, what would happen to the fund if the payments increased by 6% each year.

The matrix $\mathit{A}$ can be written as $k\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$, where $k>0$ and $0\mathrm{°}<\theta <90\mathrm{°}$.

(i) Find the value of $k$ and the value of $\theta$.

(ii) Hence describe fully the transformation represented by $\mathit{A}$.

Find the coordinates of the images under $T$ of the points $(8,0)$ and $(8,8)$.

Find $\det \mathit{A}$, and hence find the area of $S_1$.

Amara imagines continuing the pattern indefinitely. Show that the total area of all the squares $S_0,S_1,S_2,\dots$ is finite, and find this total area.

The point $C$ is the invariant point of $T$, so that $T$ maps $C$ to itself.

Find the coordinates of $C$.

Let $\mathit{c}$ be the position vector of $C$. Show that, for every position vector $\mathit{x}$, $T(\mathit{x})-\mathit{c}=\mathit{A}(\mathit{x}-\mathit{c})$.

Hence show that the distance $C{P}_n={0.75}^n\times C{P}_0$.

Amara draws the squares $S_0,S_1,S_2,\dots$ only while the vertex $P_n$ is at least 0.2 cm from $C$.

Find the number of squares in her logo.

Amara joins $P_0$ to $P_1$, $P_1$ to $P_2$, and so on, with straight line segments.

(i) Show that $P_{n+1}-P_n$, as a vector, is ${\mathit{A}}^n\left(\begin{array}{c}8\\ 0\end{array}\right)$.

(ii) Hence find the total length of this path if it were continued indefinitely.

Amara's colleague, Femi, suggests replacing $\mathit{A}$ with $\mathit{B}=\left(\begin{array}{cc}0.45& 0.6\\ 0.6& -0.45\end{array}\right)$.

The matrix $\mathit{B}$ represents a reflection in the line $y=(\tan \alpha )x$ combined with an enlargement, centred at the origin, with scale factor 0.75.

Find the equation of the line of reflection.

Find ${\mathit{B}}^2$. Hence explain why Femi's design would not produce a spiral of squares.