1.04j Sum to infinity: convergent geometric series |r|<1

The first term of a geometric series is 8. The sum to infinity of the series is 10. Find the common ratio. [3]

The first term of a geometric series is 5.4 and the common ratio is 0.1.

1. Find the fourth term of the series. [1]
2. Find the sum to infinity of the series. [2]

\includegraphics{figure_12} A branching plant has stems, nodes, leaves and buds. • There are 7 leaves at each node. • From each node, 2 new stems grow. • At the end of each final stem, there is a bud. Fig. 12 shows one such plant with 3 stages of nodes. It has 15 stems, 7 nodes, 49 leaves and 8 buds.

1. One of these plants has 10 stages of nodes.
   1. How many buds does it have? [2]
   2. How many stems does it have? [2]
2. 1. Show that the number of leaves on one of these plants with \(n\) stages of nodes is $$7(2^n - 1).$$ [2]
   2. One of these plants has \(n\) stages of nodes and more than 200000 leaves. Show that \(n\) satisfies the inequality \(n > \frac{\log_{10} 200007 - \log_{10} 7}{\log_{10} 2}\). Hence find the least possible value of \(n\). [4]

\(S\) is the sum to infinity of a geometric progression with first term \(a\) and common ratio \(r\).

1. Another geometric progression has first term \(2a\) and common ratio \(r\). Express the sum to infinity of this progression in terms of \(S\). [1]
2. A third geometric progression has first term \(a\) and common ratio \(r^2\). Express, in its simplest form, the sum to infinity of this progression in terms of \(S\) and \(r\). [2]

A hot drink when first made has a temperature which is \(65°C\) higher than room temperature. The temperature difference, \(d °C\), between the drink and its surroundings decreases by \(1.7\%\) each minute.

1. Show that 3 minutes after the drink is made, \(d = 61.7\) to 3 significant figures. [2]
2. Write down an expression for the value of \(d\) at time \(n\) minutes after the drink is made, where \(n\) is an integer. [1]
3. Show that when \(d < 3\), \(n\) must satisfy the inequality $$n > \frac{\log_{10} 3 - \log_{10} 65}{\log_{10} 0.983}.$$ Hence find the least integer value of \(n\) for which \(d < 3\). [4]
4. The temperature difference at any time \(t\) minutes after the drink is made can also be expressed as \(d = 65 \times 10^{-kt}\), for some constant \(k\). Use the value of \(d\) for 1 minute after the drink is made to calculate the value of \(k\). Hence find the temperature difference 25.3 minutes after the drink is made. [4]

The second term of a geometric progression is 24. The sum to infinity of this progression is 150. Write down two equations in \(a\) and \(r\), where \(a\) is the first term and \(r\) is the common ratio. Solve your equations to find the possible values of \(a\) and \(r\). [5]

A geometric series has common ratio \(\frac{1}{3}\). Given that the sum of the first four terms of the series is 200,

1. find the first term of the series, [3]
2. find the sum to infinity of the series. [2]

The second and fifth terms of a geometric series are \(-48\) and \(6\) respectively.

1. Find the first term and the common ratio of the series. [5]
2. Find the sum to infinity of the series. [2]
3. Show that the difference between the sum of the first \(n\) terms of the series and its sum to infinity is given by \(2^{6-n}\). [5]

A geometric series has first term 75 and second term \(-15\).

1. Find the common ratio of the series. [2]
2. Find the sum to infinity of the series. [2]

A geometric progression has third term 36 and fourth term 27. Find

1. the common ratio, [2]
2. the fifth term, [2]
3. the sum to infinity. [4]

A geometric progression has first term 75 and second term \(-15\).

1. Find the common ratio. [2]
2. Find the sum to infinity. [2]

Jill has 3 daughters and no sons. They are generation 1 of Jill's descendants. Each of her daughters has 3 daughters and no sons. Jill's 9 granddaughters are generation 2 of her descendants. Each of her granddaughters has 3 daughters and no sons; they are descendant generation 3. Jill decides to investigate what would happen if this pattern continues, with each descendant having 3 daughters and no sons.

1. How many of Jill's descendants would there be in generation 8? [2]
2. How many of Jill's descendants would there be altogether in the first 15 generations? [3]
3. After \(n\) generations, Jill would have over a million descendants altogether. Show that \(n\) satisfies the inequality $$n > \frac{\log_{10}2000003}{\log_{10}3} - 1.$$ Hence find the least possible value of \(n\). [4]
4. How many fewer descendants would Jill have altogether in 15 generations if instead of having 3 daughters, she and each subsequent descendant has 2 daughters? [3]

The second term of a geometric progression is 24. The sum to infinity of this progression is 150. Write down two equations in \(a\) and \(r\), where \(a\) is the first term and \(r\) is the common ratio. Solve your equations to find the possible values of \(a\) and \(r\). [5]

\(S\) is the sum to infinity of a geometric progression with first term \(a\) and common ratio \(r\).

1. Another geometric progression has first term \(2a\) and common ratio \(r\). Express the sum to infinity of this progression in terms of \(S\). [1]
2. A third geometric progression has first term \(a\) and common ratio \(r^2\). Express, in its simplest form, the sum to infinity of this progression in terms of \(S\) and \(r\). [2]

A geometric progression has first term \(a\) and common ratio \(r\). The second term is 6 and the sum to infinity is 25.

1. Write down two equations in \(a\) and \(r\). Show that one possible value of \(a\) is 10 and find the other possible value of \(a\). Write down the corresponding values of \(r\). [7]
2. Show that the ratio of the \(n\)th terms of the two geometric progressions found in part (i) can be written as \(2^{n-2} : 3^{n-2}\). [3]

The second term of a geometric progression is 18 and the fourth term is 2. The common ratio is positive. Find the sum to infinity of this progression. [5]

A geometric progression has 6 as its first term. Its sum to infinity is 5. Calculate its common ratio. [3]

\includegraphics{figure_8} The diagram shows the curve with equation \(y = \frac{1}{x+1}\). A set of \(n\) rectangles of unit width is drawn, starting at \(x = 0\) and ending at \(x = n\), where \(n\) is an integer.

1. By considering the areas of these rectangles, explain why $$\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n+1} < \ln(n+1).$$ [5]
2. By considering the areas of another set of rectangles, show that $$1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} > \ln(n+1).$$ [2]
3. Hence show that $$\ln(n+1) + \frac{1}{n+1} < \sum_{r=1}^{n+1} \frac{1}{r} < \ln(n+1) + 1.$$ [2]
4. State, with a reason, whether \(\sum_{r=1}^{\infty} \frac{1}{r}\) is convergent. [2]

Convergent infinite series \(C\) and \(S\) are defined by \begin{align} C &= 1 + \frac{1}{4} \cos \theta + \frac{1}{4} \cos 2\theta + \frac{1}{8} \cos 3\theta + \ldots,
S &= \frac{1}{2} \sin \theta + \frac{1}{4} \sin 2\theta + \frac{1}{8} \sin 3\theta + \ldots. \end{align}

1. Show that \(C + iS = \frac{2}{2 - e^{i\theta}}\). [4]
2. Hence show that \(C = \frac{4 - 2\cos \theta}{5 - 4\cos \theta}\) and find a similar expression for \(S\). [4]

In this question you should assume that \(-1 < x < 1\).

1. For the binomial expansion of \((1 - x)^{-2}\)
   1. find and simplify the first four terms, [2]
   2. write down the term in \(x^n\). [1]
2. Write down the sum to infinity of the series \(1 + x + x^2 + x^3 + \ldots\). [1]
3. Hence or otherwise find and simplify an expression for \(2 + 3x + 4x^2 + 5x^3 + \ldots\) in the form \(\frac{a - x}{(b - x)^2}\) where \(a\) and \(b\) are constants to be determined. [3]

The positive integers \(x\), \(y\) and \(z\) are the first, second and third terms, respectively, of an arithmetic progression with common difference \(-4\). Also, \(x\), \(\frac{15}{y}\) and \(z\) are the first, second and third terms, respectively, of a geometric progression.

1. Show that \(y\) satisfies the equation \(y^4 - 16y^2 - 225 = 0\). [4]
2. Hence determine the sum to infinity of the geometric progression. [4]

On the first day of each month, Kate pays £50 into a savings account. Interest is paid on the total amount in the account on the last day of each month. The interest rate is 0.2% At the end of the \(n\)th month, the total amount of money in Kate's savings account is £\(T_n\) Kate correctly calculates \(T_1\) and \(T_2\) as shown below: \(T_1 = 50 \times 1.002 = 50.10\) \(T_2 = (T_1 + 50) \times 1.002\) \(= ((50 \times 1.002) + 50) \times 1.002\) \(= 50 \times 1.002^2 + 50 \times 1.002\) \(\approx 100.30\)

1. Show that \(T_3\) is given by $$T_3 = 50 \times 1.002^3 + 50 \times 1.002^2 + 50 \times 1.002$$ [1 mark]
2. Kate uses her method to correctly calculate how much money she can expect to have in her savings account at the end of 10 years.
   1. Find the amount of money Kate expects to have in her savings account at the end of 10 years. [3 marks]
   2. The amount of money in Kate's savings account at the end of 10 years may not be the amount she has correctly calculated. Explain why. [1 mark]