Sequences and Series

Sequences and Series

Sequence is a list of numbers with a predefined rule or formula.

The numbers are called as terms and we can find the pattern in the sequence.

There are quite a limitless number of possible sequences can be made, we however will mainly focus on two main categories of sequences, the arithmetic sequence and the geometric sequence.

Arithmetic Sequence

In arithmetic sequence or progression any two consecutive terms always have the same difference.

\(\\[12pt]\hspace{2em}{\small {u}_{n} \ = \ a \ + \ (n \ – \ 1) d }\)
\(\\[7pt]{\small {u}_{n} }\) is the \({\small \boldsymbol{{n}^{\textrm{th}}} }\) term
\(\\[7pt]{\small a }\) is the first term or \({\small \boldsymbol{{u}_{1}} }\)
\({\small d }\) is the common difference.

Example:

\(\\[7pt]{\small a \ \ \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 }\)
\(\\[7pt]{\small {u}_{2} \ = \ 15, \ {u}_{3} \ = \ 20, \ {u}_{4} \ = \ 25, \ \textrm{etc} }\)
\(\\[7pt]{\small d \ \ \ = \ 5}\)
\(\\[7pt]{\small {u}_{n} \ = \ a \ + \ (n \ – \ 1) d }\)
\(\\[7pt]{\small {u}_{n} \ = \ 10 \ + \ (n \ – \ 1) \ \times \ 5 }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 5n \ – \ 5 }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 5n \ + \ 5 }\)

So, for example, if we want to find the 10th term of the sequence above,

\(\\[7pt]{\small {u}_{10} \ = \ 5(10) \ + \ 5 }\)
\(\hspace{1.6em}{\small = \ 55 }\)

Geometric Sequence

In geometric sequence or progression any two consecutive terms always have the same ratio.

\(\\[12pt]\hspace{2em}{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }\)
\(\\[7pt]{\small {u}_{n} }\) is the \({\small \boldsymbol{{n}^{\textrm{th}}} }\) term
\(\\[7pt]{\small a }\) is the first term or \({\small \boldsymbol{{u}_{1}} }\)
\({\small r }\) is the common ratio.

Example:

\(\\[7pt]{\small a \ \ \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 }\)
\(\\[7pt]{\small {u}_{2} \ = \ 50, \ {u}_{3} \ = \ 25, \ {u}_{4} \ = \ 12.5, \ \textrm{etc} }\)
\(\\[12pt]{\small r \ \ \ = \ {\large\frac{1}{2}} }\)
\(\\[7pt]{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }\)
\(\\[12pt]{\small {u}_{n} \ = \ 100 \ \times \ { \big({\large\frac{1}{2}} \big) }^{ (n \ – \ 1) } }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 \ \times \ {2}^{(-n+1)} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 200 \ \times \ {2}^{-n} }\)

So, for example, if we want to find the 10th term of the sequence above,

\(\\[7pt]{\small {u}_{10} \ = \ 200 \ \times \ {2}^{-10} }\)
\(\hspace{1.6em}{\small = \ 0.1953125 }\)

Meanwhile, we can also calculate the sum of n terms, that is, the addition from the first term of the sequence up to the \({\small {n}^{\textrm{th}} }\) term.

The addition of the sequence is called series.

The formula for both arithmetic series and geometric series are as follows:

Arithmetic Series

\(\\[18pt]\hspace{1.5em}{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ ( a \ + \ {u}_{n} ) \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] }\)
\(\\[7pt]{\small {S}_{n} }\) is the sum up to \({\small \boldsymbol{{n}^{\textrm{th}}} }\) term.

Example:

\(\\[7pt]{\small {S}_{1} \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 }\)
\(\\[7pt]{\small {S}_{2} \ = \ {u}_{1} \ + \ {u}_{2} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 }\)
\(\\[7pt]{\small {S}_{3} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 \ + \ 20 }\)
\(\\[7pt]{\small {S}_{4} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} \ + \ {u}_{4} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 \ + \ 20 \ + \ 25}\)
\(\\[12pt]\), etc
\(\\[7pt]{\small d \ \ \ = \ 5}\)
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] } \)
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2(10) \ + \ (n \ – \ 1) \ \times \ 5 \big] } \)
\(\\[12pt]\hspace{1.4em}{\small = \ {\large\frac{n}{2}} \ ( 20 \ + \ 5n \ – \ 5 ) } \)
\(\\[7pt]\hspace{1.4em}{\small = \ {\large\frac{5}{2}}{n}^2 \ + \ {\large\frac{15}{2}}n }\)

So, for example, if we want to find the sum of the first 10 terms of the sequence above,

\(\\[12pt]{\small {S}_{10} \ = \ {\large\frac{5}{2}}{(10)}^2 \ + \ {\large\frac{15}{2}}n }\)
\(\hspace{1.6em}{\small = \ 325 }\)

Geometric Series

\(\\[12pt]\hspace{2em}{\small {S}_{n} \ = \ {\large\frac{a \ ( {r}^{ n } \ – \ 1 ) }{r \ – \ 1} } }\) for \({\small |r| \ \gt \ 1 }\)
\(\\[7pt]\) or,
\(\\[20pt]\hspace{2em}{\small {S}_{n} \ = \ {\large\frac{a \ ( 1 \ – \ {r}^{ n } )}{1 \ – \ r } } }\) for \({\small |r| \ \lt \ 1 }\)
\(\\[7pt]{\small {S}_{n} }\) is the sum up to \({\small \boldsymbol{{n}^{\textrm{th}}} }\) term.

Example:

\(\\[7pt]{\small {S}_{1} \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 }\)
\(\\[7pt]{\small {S}_{2} \ = \ {u}_{1} \ + \ {u}_{2} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 }\)
\(\\[7pt]{\small {S}_{3} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 \ + \ 25 }\)
\(\\[7pt]{\small {S}_{4} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} \ + \ {u}_{4} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 \ + \ 25 \ + \ 12.5}\)
\(\\[12pt]\), etc
\(\\[12pt]{\small r \ \ \ = \ {\large\frac{1}{2}} }\)
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{a \ ( 1 \ – \ {r}^{ n } )}{1 \ – \ r } } }\)
\(\\[18pt]{\small {S}_{n} \ = \ {\large\frac{100 \ \big( 1 \ – \ {{\large (\frac{1}{2} )} }^{ n } \big)}{1 \ – \ {\large\frac{1}{2}} } } }\)
\(\\[18pt]{\small {S}_{n} \ = \ {\large\frac{100 \ \big( 1 \ – \ {2} ^{ -n } \big)}{ {\large\frac{1}{2}} } } }\)
\(\\[10pt]{\small {S}_{n} \ = \ 200 \ \times \ \big( 1 \ – \ {2 }^{ -n } \big) }\)

So, for example, if we want to find the sum of the first 10 terms of the sequence above,

\(\\[10pt]{\small {S}_{n} \ = \ 200 \ \times \ \big( 1 \ – \ {(2) }^{ -10 } \big) }\)
\(\hspace{1.4em}{\small = \ 199.8046875 }\)

In geometric series, we can also find the sum to infinity given that the series is convergent \({\small (|r| \ \lt \ 1) }\).

Since \({\small \displaystyle \lim_{n \to \infty} 1 \ – \ {r}^{ n } \ = \ 1 \ }\) for \({\small \ (|r| \ \lt \ 1) }\),

\(\\[12pt]\hspace{2em}{\small {S}_{\infty} \ = \ {\large\frac{a}{1 \ – \ r } } }\)
\(\\[7pt]{\small {S}_{\infty} }\) is the sum to infinity.

Example:

\(\\[18pt]{\small {S}_{\infty} \ = \ {\large\frac{ a }{1 \ – \ r } } }\)
\(\\[18pt]{\small {S}_{\infty} \ = \ {\large\frac{ 100 }{1 \ – \ {\large\frac{1}{2}} } } }\)
\(\hspace{1.6em}{\small = \ 200 }\)

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)

EXAMPLE:

\({\small 1.\enspace}\) A runner who is training for a long-distance race plans to run increasing distances each day for 21 days. She will run x km on day 1, and on each subsequent day she will increase the distance by 10% of the previous day’s distance. On day 21 she will run 20 km.
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Find the distance she must run on day 1 in order to achieve this. Give your answer in km correct to 1 decimal place.
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Find the total distance she runs over the 21 days.

Show solution:

\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) We are attempting to tackle a geometric sequence question.
\(\\[1pt]\)
There are 21 terms in our sequence.
\(\\[1pt]\)
The ratio of the geometric sequence is 1.1. The value of the next term is increased in a 1.1 ratio of the current term.
\(\\[1pt]\)
The value of the 21-th term is also given.
\(\\[1pt]\)
Find the value of \({\small x }\).
\(\\[1pt]\)
\(\\[7pt]{\small {u}_{1} \ = \ x }\)
\(\\[7pt]{\small n \ \ \ = \ 21 }\)
\(\\[7pt]{\small r \ \ \ = \ 1.1 }\)
\({\small {u}_{21} = \ 20 }\)
\(\\[1pt]\)
We can use the \({\small {u}_{n} }\) formula to solve for the unknown value \({\small x }\).
\(\\[1pt]\)
\(\\[7pt]{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }\)
\(\\[7pt]{\small {u}_{21} \ = \ x \ \times \ {1.1}^{ (21 \ – \ 1) } }\)
\(\\[7pt]{\small \ 20 \ = \ x \ \times \ {1.1}^{ 20 } }\)
\(\\[15pt]{\small x \ \ \ = \ { \large\frac{20}{ {1.1}^{ 20 } } } }\)
\(\\[10pt]\hspace{1.4em}{\small \approx \ 3.0 }\)
Hence, she must run 3.0 km on day 1.
\(\\[1pt]\)
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Find \({\small {S}_{n}}\) for \({\small n \ = \ 21 }\).
\(\\[1pt]\)
\(\\[15pt]{\small {S}_{n} \ \ = \ {\large\frac{a \ ( {r}^{ n } \ – \ 1 )}{ r \ – \ 1 } } }\)
\(\\[15pt]{\small {S}_{21} \ = \ {\large\frac{3.0 \ \times \ ( {1.1}^{ 21 } \ – \ 1 )}{ 1.1 \ – \ 1 } } }\)
\(\\[10pt]{\small \hspace{1.2em} \ \approx \ 192 }\)
Therefore, the total distance she runs over 21 days is 192 km.

\(\\[1pt]\)
\({\small 2.\enspace}\) The first, second and third terms of a geometric progression are x, x – 3 and x – 5 respectively.
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Find the value of x.
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Find the fourth term of the progression.
\({\small\hspace{2.8em}(\textrm{iii}).\hspace{0.5em}}\) Find the sum to infinity of the progression.

Show solution:

\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) The ratio between any two consecutive terms is always the same.
\(\\[1pt]\)
Thus,
\(\\[1pt]\)
\(\\[15pt]\hspace{2.1em}{\small {\large\frac{ {u}_{2} }{ {u}_{1} } } \ = \ {\large\frac{ {u}_{3} }{ {u}_{2} } } }\)
\(\\[15pt]\hspace{1.2em}{\small {\large\frac{ x \ – \ 3 }{ x }} \ = \ {\large\frac{ x \ – \ 5 }{ x \ – \ 3 } }}\)
Cross multiply both side of the equations,
\(\\[1pt]\)
\(\\[7pt]\hspace{1.7em}{\small {(x \ – \ 3)}^{2} \ = \ x \ \times \ (x \ – \ 5) }\)
\(\\[7pt]{\small {x}^{2} \ – \ 6x \ + \ 9 \ = \ {x}^{2} \ – \ 5x }\)
\(\hspace{4.4em} {\small x = \ 9 }\)
\(\\[1pt]\)
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) We can calculate the common ratio first.
\(\\[1pt]\)
\(\\[15pt]{\small r \ = \ {\large\frac{ x \ – \ 3 }{ x } } }\)
\(\\[15pt]{\small r \ = \ {\large\frac{ 9 \ – \ 3 }{ 9 } } }\)
\(\\[7pt]{\small r \ = \ {\large\frac{ 2 }{ 3 } } }\)
\(\\[1pt]\)
Then, the fourth term is,
\(\\[1pt]\)
\(\\[7pt]{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }\)
\(\\[15pt]{\small {u}_{4} \ = \ x \ \times \ { \big( {\large\frac{ 2 }{ 3 } } \big) }^{ (4 \ – \ 1) } }\)
\(\\[15pt]\hspace{0.4em}{\small \ \ \ = \ 9 \ \times \ { \big ({\large\frac{ 2 }{ 3 } } \big) }^{ 3 } }\)
\(\\[10pt]\hspace{0.4em}{\small \ \ \ = \ {\large\frac{ 8 }{ 3 } } }\) or 2.67
\(\\[1pt]\)
\({\small\hspace{2.8em}(\textrm{iii}).\hspace{0.5em}}\) The sum to infinity is,
\(\\[1pt]\)
\(\\[18pt]{\small {S}_{\infty} \ = \ {\large\frac{ a }{1 \ – \ r } } }\)
\(\\[18pt]{\small {S}_{\infty} \ = \ {\large\frac{ 9 }{1 \ – \ {\large\frac{2}{3}} } } }\)
\(\hspace{1.6em}{\small = \ 27 }\)

\(\\[1pt]\)
\({\small 3.\enspace}\) In an arithmetic progression, the sum of the first ten terms is equal to the sum of the next five terms. The first term is a.
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Show that the common difference of the progression is \({\small {\large \frac{1}{3} } a}\).
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Given that the tenth term is 36 more than the fourth term, find the value of a.

Show solution:

\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Show that \({\small d \ = \ {\large \frac{1}{3} } a \ }\).
\(\\[1pt]\)
“the sum of the first ten terms” is \({\small {S}_{10} }\)
\(\\[1pt]\)
“the sum of the next five terms” is,
\(\\[1pt]\)
\(\\[7pt]{\small \hspace{1.2em} {u}_{11} \ + \ {u}_{12} \ + \ {u}_{13} \ + \ {u}_{14} \ + \ {u}_{15} }\)
\(\\[1pt]\)
Since,
\(\\[1pt]\)
\(\\[7pt]{\small {S}_{15} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ … \ + \ {u}_{15} }\)
\(\\[12pt]{\small {S}_{15} \ = \ {S}_{10} \ + \ {u}_{11} \ + \ {u}_{12} \ + \ … \ + \ {u}_{15} }\)
\(\\[15pt]{\small {u}_{11} \ + \ {u}_{12} \ + \ … \ + \ {u}_{15} \ = \ {S}_{15} \ – \ {S}_{10} }\)
Therefore, we can write the given statement in the question as,
\(\\[1pt]\)
\(\\[7pt]\hspace{2.1em}{\small {S}_{10} \ = \ {S}_{15} \ – \ {S}_{10} }\)
\(\\[15pt]\hspace{1.5em}{\small 2 \ {S}_{10} \ = \ {S}_{15} }\)
\(\\[15pt]\) Simplify \({\small {S}_{10} }\) and \({\small {S}_{15} }\),
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] } \)
\(\\[12pt]{\small {S}_{10} \ = \ {\large\frac{10}{2}} \ \big[ 2a \ + \ (10 \ – \ 1) \ \times \ d \big] } \)
\(\\[12pt]\hspace{1.6em}{\small = \ 5 \ ( 2a \ + \ 9d ) } \)
\(\\[17pt]\hspace{1.6em}{\small = \ 10a \ + \ 45d } \)
\(\\[12pt]{\small {S}_{15} \ = \ {\large\frac{15}{2}} \ \big[ 2a \ + \ (15 \ – \ 1) \ \times \ d \big] } \)
\(\\[12pt]\hspace{1.6em}{\small = \ {\large\frac{15}{2}} \ ( 2a \ + \ 14d ) } \)
\(\\[12pt]\hspace{1.6em}{\small = \ 15a \ + \ 105d } \)
Then,
\(\\[1pt]\)
\(\\[7pt]\hspace{3.5em} {\small 2 \ {S}_{10} \ = \ {S}_{15} }\)
\(\\[7pt]{\small 2 \ (10a \ + \ 45d) \ = \ 15a \ + \ 105d } \)
\(\\[7pt]\hspace{1.2em}{\small 20a \ + \ 90d \ = \ 15a \ + \ 105d } \)
\(\\[7pt]\hspace{4em}{\small 15d \ = \ 5a } \)
\(\\[12pt]\hspace{4.8em}{\small d \ = \ {\large \frac{1}{3} } a \enspace }\) (Q.E.D.)
\(\\[1pt]\)
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Find the first term a.
\(\\[1pt]\)
\(\\[7pt]\hspace{2.8em}{\small {u}_{10} \ = \ {u}_{4} \ + \ 36 }\)
\(\\[7pt]\hspace{1.2em}{\small a \ + \ 9d \ = \ a \ + \ 3d \ + \ 36 } \)
\(\\[15pt]\hspace{3.1em}{\small 6d \ = \ 36 } \)
\(\\[15pt]\hspace{1.2em}\) Since \({\small d \ = \ {\large \frac{1}{3} } a }\),
\(\\[12pt]\hspace{1.2em}{\small 6 \ \times \ \big({\large \frac{1}{3} } a \big) \ = \ 36 } \)
\(\\[7pt]\hspace{4.5em}{\small 2a \ = \ 36 } \)
\(\\[7pt]\hspace{4.9em}{\small a \ = \ 18 } \)

\(\\[1pt]\)
\({\small 4.\enspace}\) The sum to infinity of a geometric progression is 9 times the sum of the first four terms. Given that the first term is 12, find the value of the fifth term.

Show solution:

\(\\[10pt]\hspace{1.3em}{\small {S}_{\infty} \ = \ 9 \ \times \ {S}_{4} }\)
\(\\[18pt]{\small \ \ {\large\frac{ a }{1 \ – \ r } } \ = \ {\large\frac{9 a \ ( 1 \ – \ {r}^{ 4 } )}{1 \ – \ r } } }\)
\(\\[18pt]\require{cancel} {\small {\large\frac{ \cancel{a} }{\bcancel{1 \ – \ r} } } \ = \ {\large\frac{9 \cancel{a} \ ( 1 \ – \ {r}^{ 4 } )}{\bcancel{1 \ – \ r} } } }\)
\(\\[15pt]\hspace{1.7em}{\small {\large\frac{ 1 }{ 9 } } \ = \ 1 \ – \ {r}^{ 4 } }\)
\(\\[15pt]\hspace{1.8em}{\small {r}^{ 4 } \ = \ 1 \ – \ {\large\frac{ 1 }{ 9 } } }\)
\(\\[15pt]\hspace{1.8em}{\small {r}^{ 4 } \ = \ {\large\frac{ 8 }{ 9 } } }\)
\(\\[12pt]\) Then,
\(\\[7pt]{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }\)
\(\\[7pt]{\small {u}_{5} \ = \ a {r}^{ (5 \ – \ 1) } }\)
\(\\[7pt]{\small {u}_{5} \ = \ a {r}^{ 4 } }\)
\(\\[15pt]\hspace{1.2em}{\small = \ 12 \ \times \ {\large\frac{ 8 }{ 9 } } }\)
\(\hspace{1.2em}{\small = \ {\large\frac{ 32 }{ 3 } } \ \ }\) or 10.7

\(\\[1pt]\)
\({\small 5.\enspace}\) Two heavyweight boxers decide that they would be more successful if they competed in a lower weight class. For each boxer this would require a total weight loss of 13 kg. At the end of week 1 they have each recorded a weight loss of 1 kg and they both find that in each of the following weeks their weight loss is slightly less than the week before.
\(\\[1pt]\)
Boxer A’s weight loss in week 2 is 0.98 kg. It is given that his weekly weight loss follows an arithmetic progression.
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Write down an expression for his total weight loss after x weeks.
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) He reaches his 13 kg target during week n. Use your answer to part (i) to find the value of n.
\(\\[1pt]\)
Boxer B’s weight loss in week 2 is 0.92 kg and it is given that his weekly weight loss follows a geometric progression.
\({\small\hspace{2.8em}(\textrm{iii}).\hspace{0.5em}}\) Calculate his total weight loss after 20 weeks and show that he can never reach his target.

Show solution:

\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Find the \({\small {S}_{n} }\) after x weeks.
\(\\[1pt]\)
\(\\[7pt]{\small a \ \ \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 1 }\)
\(\\[7pt]{\small {u}_{2} \ = \ 0.98}\)
\(\\[7pt]{\small d \ \ \ = \ {u}_{2} \ – \ {u}_{1}}\)
\(\\[7pt]\hspace{1.4em}{\small = \ 0.98 \ – \ 1 }\)
\(\\[12pt]\hspace{1.4em}{\small = \ -0.02 }\)
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] } \)
With \(\\[12pt]{\small n \ = \ x }\),
\(\\[12pt]{\small {S}_{x} \ = \ {\large\frac{x}{2}} \ \big[ 2(1) \ + \ (x \ – \ 1) \ \times \ -0.02 \big] } \)
\(\\[10pt]\hspace{1.4em}{\small = \ {\large\frac{x}{2}} \ ( 2 \ – \ 0.02x \ + \ 0.02 ) } \)
\(\\[7pt]\hspace{1.4em}{\small = \ -0.01{x}^2 \ + \ 1.01x }\)
\(\\[1pt]\)
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Find n for \({\small {S}_{n} \ = \ 13 }\)
\(\\[1pt]\)
\(\\[10pt]\hspace{1.3em}{\small {S}_{x} \ = \ -0.01{x}^2 \ + \ 1.01x }\)
\(\\[10pt]\hspace{1.2em}{\small 13 \ \ = \ -0.01{n}^2 \ + \ 1.01n }\)
\(\\[10pt]\hspace{1.2em}{\small 0.01{n}^2 \ – \ 1.01n \ + \ 13 \ \ = \ 0 }\)
\(\\[25pt]\hspace{2.1em}{\small {n}^2 \ – \ 101n \ + \ 1300 \ \ = \ 0 }\)
Use the general formula for the quadratic equation,
\(\\[1pt]\)
\(\hspace{1.2em}\) For \({\small a{x}^{2} \ + \ bx \ + \ c \ = \ 0}\),
\(\\[1pt]\)
\(\hspace{2em} {\small x \ = \ {\large \frac{-b \ \pm \ \sqrt{ {b}^{2} \ – \ 4ac }}{2a} } } \)
\(\\[1pt]\)
Therefore,
\(\\[1pt]\)
\(\hspace{1.2em}\)\({\small {n}^2 \ – \ 101n \ + \ 1300 \ \ = \ 0 }\)
\(\\[1pt]\)
\(\hspace{2em} {\small n \ = \ {\large \frac{101 \ \pm \ \sqrt{ {(-101)}^{2} \ – \ 4(1)(1300) }}{2(1)} } }\)
\(\\[1pt]\)
\(\hspace{2em} {\small n \ = \ {\large \frac{101 \ \pm \ \sqrt{ 5001 }}{2} } }\)
\(\\[1pt]\)
\(\hspace{2em}{\small n \ \approx \ 15.1 \ }\) or \( \ {\small n \ \approx \ 85.9 }\)
\(\\[1pt]\)
Hence, \({\small n \ = \ 16 }\). Note that \( \ {\small n \ = \ 86 \ }\) is ignored since we are interested in the smaller value of n.
\(\\[1pt]\)
\(\\[1pt]\)
\({\small\hspace{2.8em}(\textrm{iii}).\hspace{0.7em}}\) Show that \({\small {S}_{20} \ \lt \ 13 }\).
\(\\[1pt]\)
\(\\[7pt]{\small a \ \ \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 1 }\)
\(\\[7pt]{\small {u}_{2} \ = \ 0.92}\)
\(\\[15pt]{\small r \ \ \ = \ {\large\frac{ {u}_{2} }{ {u}_{1} } } }\)
\(\\[15pt]\hspace{1.4em}{\small = {\large\frac{ 0.92 }{ 1 } } }\)
\(\\[12pt]\hspace{1.4em}{\small = \ 0.92 }\)
\(\\[18pt]{\small {S}_{n} \ = \ {\large\frac{a \ ( 1 \ – \ {r}^{ n } )}{1 \ – \ r } } }\)
With \(\\[12pt]{\small n \ = \ 20 }\),
\(\\[15pt]{\small {S}_{20} \ = \ {\large\frac{1 \ \times \ ( 1 \ – \ {0.92}^{ 20 } )}{1 \ – \ 0.92 } } }\)
\(\\[15pt]\hspace{1.7em}{\small = \ {\large\frac{ 1 \ – \ {0.92}^{ 20 } }{0.08 } } }\)
\(\hspace{1.7em}{\small \approx \ 10.1 \ \lt \ 13 \ }\) (Q.E.D)

\(\\[1pt]\)
\({\small 6.\enspace}\) The sum of the first twenty terms of an arithmetic progression is 50, and the sum of the next twenty terms is -50. Find the sum of the first hundred terms of the progression.

Show solution:

“the sum of the first twenty terms” is \({\small {S}_{20} }\)
\(\\[1pt]\)
“the sum of the next twenty terms” is,
\(\\[1pt]\)
\(\\[7pt]{\small \hspace{1.2em} {u}_{21} \ + \ {u}_{22} \ + \ … \ + \ {u}_{40} }\)
\(\\[1pt]\)
Since,
\(\\[1pt]\)
\(\\[7pt]{\small {S}_{40} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ … \ + \ {u}_{40} }\)
\(\\[12pt]{\small {S}_{40} \ = \ {S}_{20} \ + \ {u}_{21} \ + \ {u}_{22} \ + \ … \ + \ {u}_{40} }\)
\(\\[15pt]{\small {u}_{21} \ + \ {u}_{22} \ + \ … \ + \ {u}_{40} \ = \ {S}_{40} \ – \ {S}_{20} }\)
\(\\[15pt]\) Simplify \({\small {S}_{20} }\) and \({\small {S}_{40} }\),
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] } \)
\(\\[12pt]{\small {S}_{20} \ = \ {\large\frac{20}{2}} \ \big[ 2a \ + \ (20 \ – \ 1) \ \times \ d \big] } \)
\(\\[12pt]{\small \hspace{0.2 em} 50 \ = \ 10 \ ( 2a \ + \ 19d ) } \)
\(\\[17pt]{\small \hspace{0.6 em} 5 \ = \ 2a \ + \ 19d } \quad \ldots \quad {\small (1)} \)
\(\\[12pt]{\small {S}_{40} \ = \ {\large\frac{40}{2}} \ \big[ 2a \ + \ (40 \ – \ 1) \ \times \ d \big] } \)
\(\\[12pt]{\small \hspace{0.6 em} 0 \ = \ 20 \ ( 2a \ + \ 39d ) } \)
\(\\[15pt]{\small \hspace{0.6 em} 0 \ = \ 2a \ + \ 39d } \quad \ldots \quad {\small (2)} \)
Then, by using substitution or elimination method we can find a and d from (1) and (2),
\(\\[1pt]\)
\(\\[10pt] \hspace{2 em} {\small a \ = \ {\large\frac{39}{8} } }\)
\(\\[20pt] \hspace{2 em} {\small d \ = \ {\large – \frac{1}{4} } }\)
\(\\[12pt]\) \({\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] } \)
\(\\[18pt]\) \({\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \Big[ 2 \big({\large\frac{39}{8} }\big) \ + \ (n \ – \ 1) \ \big({\large – \frac{1}{4} }\big) \Big] } \)
\(\\[18pt]\) \({\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \Big[ {\large\frac{39}{4} } \ – \ {\large \frac{1}{4} }n \ + \ {\large \frac{1}{4} } \Big] } \)
\(\\[18pt]\) \({\small \hspace{1.4em} = \ {\large – \frac{1}{8} }{n}^{2} \ + \ 5n }\)
Therefore,
\(\\[1pt]\)
\(\\[15pt]\) \({\small {S}_{100} \ = \ {\large – \frac{1}{8} }{(100)}^{2} \ + \ 5(100) }\)
\({\small \hspace{2.2em} = \ -750 }\)

\(\\[1pt]\)
\({\small 7.\enspace}\) In 1971 a newly-built flat was sold with a 999-year lease. The terms of the sale included a requirement to pay ‘ground rent’ yearly. The ground rent was set at £28 per year for the first 21 years of the lease, increasing by £14 to £42 per year for the next 21 years, and then increasing again by £14 at the end of each subsequent period of 21 years.
\({\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em}}\) Find how many complete 21-year periods there would be if the lease ran for the full 999 years, and how many years there would be left over.
\({\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}\) Find the total amount of ground rent that would be paid in all of the complete 21-year periods of the lease.

Show solution:

\(\\[12pt]{\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em} \ {\large\frac{999}{21} } \ = \ 47 \ \textrm{R} \ 12 }\)
So, there will be 47 complete 21-year periods of the lease and 12 years left over.
\(\\[1pt]\)
\({\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}\) We can construct the arithmetic progression of ‘ground rent’ paid yearly for each of 21-year periods.
\(\\[1pt]\)

\(\\[1pt]\)
The above arithmetic progression can then be rewritten as ‘ground rent’ paid per 21-year periods.
\(\\[1pt]\)

\(\\[1pt]\)
\(\\[7pt]{\small a \ \ \ = \ 588 }\)
\(\\[7pt]{\small d \ \ \ = \ 294}\)
\(\\[15pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] } \)
\(\\[10pt]\) There would be 47 periods,
\(\\[10pt]\) So, for \({\small n \ = \ 47 }\),
\(\\[15pt]{\small {S}_{47} \ = \ {\large\frac{47}{2}} \ \big[ 2(588) \ + \ (47 \ – \ 1) \ \times \ 294 \big] } \)
\(\\[15pt]\hspace{1.6em}{\small = \ {\large\frac{47}{2}} \ ( 1176 \ + \ 13524 ) } \)
\(\\[15pt]\hspace{1.6em}{\small = \ {\large\frac{47}{2}} \ \times \ 14700 }\)
\(\\[12pt]\hspace{1.6em}{\small = \ 345 \ 450 }\)
Hence, the total amount of ground rent that would be paid is £345 450.

\(\\[1pt]\)
\({\small 8.\enspace}\) Jayesh invests $100 in a savings acoount on the first day of each month for one complete year. The account pays interest at \( {\small \frac{1}{2} \% \ }\) for each complete month. How much does Jayesh have invested at the end of the year (but before making a thirteenth payment)?

Show solution:

So, Jayesh invests $100 each month from January to December (12 months).
\(\\[1pt]\)
\(\\[10pt]\) For $100 he invests in January,
\(\\[7pt]\hspace{0.5em}\) – at the end of Jan it’ll worth:
\(\\[7pt]\hspace{1.5em} {\small $100 \ + \ \frac{1}{2} \% \times $100 }\)
\(\\[7pt]\hspace{1.5em} {\small $100 \times (1 + \frac{1}{2} \%) }\)
\(\\[12pt]\hspace{1.5em} {\small $100 \times 1.005 }\)
\(\\[7pt]\hspace{0.5em}\) – at the end of Feb it’ll worth:
\(\\[7pt]\hspace{1.5em} {\small $100 \times 1.005 \ + \ \frac{1}{2} \% \times $100 \times 1.005 }\)
\(\\[7pt]\hspace{1.5em} {\small $100 \times 1.005 \times (1 + \frac{1}{2} \%) }\)
\(\\[7pt]\hspace{1.5em} {\small $100 \times 1.005 \times 1.005 }\)
\(\\[7pt]\hspace{1.5em} {\small $100 \times {1.005}^{2} }\)
\(\\[7pt] \hspace{4.5em} {\large \vdots} \)
\(\\[7pt]\hspace{0.5em}\) – at the end of Dec it’ll worth:
\(\\[7pt]\hspace{1.5em} {\small $100 \times {1.005}^{12} }\)
\(\\[1pt]\)
\(\\[10pt]\) For $100 he invests in February,
\(\\[7pt]\hspace{0.5em}\) – at the end of Feb it’ll worth:
\(\\[12pt]\hspace{1.5em} {\small $100 \times 1.005 }\)
\(\\[7pt]\hspace{0.5em}\) – at the end of March it’ll worth:
\(\\[7pt]\hspace{1.5em} {\small $100 \times {1.005}^{2} }\)
\(\\[7pt] \hspace{4.5em} {\large \vdots} \)
\(\\[7pt]\hspace{0.5em}\) – at the end of Dec it’ll worth:
\(\\[7pt]\hspace{1.5em} {\small $100 \times {1.005}^{11} }\)
\(\\[1pt]\)
Therefore, for every $100 Jayesh invests from January till December, the total amount he has invested, $S,
\(\\[1pt]\)
\({\scriptsize S = 100 \times 1.005 + 100 \times {1.005}^{2} + \ \cdots \ + 100 \times {1.005}^{12} }\)
\(\\[1pt]\)
\(\\[7pt]\) This is a geometric series with,
\(\\[10pt] \hspace{1.2em} {\small a \ = \ 100 \times 1.005 }\)
\(\\[10pt] \hspace{1.2em} {\small r \ = \ 1.005 }\)
\(\\[10pt] \hspace{1.2em} {\small n \ = \ 12 }\)
\(\\[18pt]{\small {S}_{n} \ = \ {\large\frac{a \ ( {r}^{ n } \ – \ 1)}{ r \ – \ 1 } } }\)
\(\\[15pt]{\small {S}_{12} \ = \ {\large\frac{(100 \times 1.005) \ \times \ ( {1.005}^{ 12 } \ – \ 1 )}{1.005 \ – \ 1 } } }\)
\(\\[12pt]\hspace{1.7em}{\small \approx \ 1 \ 239.72 }\)
Hence, Jayesh has invested $1 239.72 at the end of the year.

\(\\[1pt]\)
\({\small 9.\enspace}\) Neeta takes out a 25-year mortgage of $40 000 to buy her house. Compound interest is charged on the loan at a rate of 8% per annum. She has to pay off the mortgage with 25 equal payments, the first of which is to be one year after the loan is taken out. Continue the following argument to calculate the value of each annual payment.
\(\\[1pt]\)
\({\small \hspace{1.2em} \bullet \hspace{1.2em} }\) After 1 year she owes \({\small $(40 000 \times 1.08) }\) (loan plus interest) less the payment made, $P, that is, she owes \({\small $(40 000 \times 1.08 \ – P)}\).
\(\\[1pt]\)
\( {\small \hspace{1.2em} \bullet \hspace{1.2em} }\) After 2 years she owes \({\small $\big((40 000 \times 1.08 \ – P) \times 1.08 \ – P \big) }\).
\(\\[1pt]\)
\({\small \hspace{1.2em} \bullet \hspace{1.2em} }\) After 3 years she owes \({\small $\Big(\big((40 000 \times 1.08 \ – P) \times 1.08 \ – P \big) \times 1.08 \ – P \Big)}\).
\(\\[1pt]\)
At the end of 25 years this (continued) expression must be zero. Form an equation in P and solve it.

Show solution:

\(\\[10pt]\hspace{0.5em}\) – After 1 year she owes:
\(\\[7pt]{\small {u}_{1} = \ (40 000 \times 1.08 \ – P)}\)
\(\\[1pt]\)
\(\\[10pt]\hspace{0.5em}\) – After 2 years she owes:
\(\\[7pt]{\small {u}_{2} = \ ((40 000 \times 1.08 \ – P) \times 1.08 \ – P ) }\)
\(\\[7pt]{\small \hspace{1.1em} = \ 40 000 \times {1.08}^{2} \ – 1.08P \ – P }\)
\(\\[7pt]{\small \hspace{1.1em} = \ 40 000 \times {1.08}^{2} \ – P(1 + 1.08) }\)
\(\\[1pt]\)
\(\\[10pt]\hspace{0.5em}\) – After 3 years she owes:
\(\\[7pt]{\small {u}_{3} = \ (((40 000 \times 1.08 \ – P) \times 1.08 \ – P ) \times 1.08 \ – P ) }\)
\(\\[7pt]{\small \hspace{1.1em} = \ (40 000 \times {1.08}^{2} \ – P(1 + 1.08)) \times 1.08 \ – P }\)
\(\\[7pt]{\small \hspace{1.1em} = \ 40 000 \times {1.08}^{3} \ – P(1 + 1.08+{1.08}^{2}) }\)
\(\\[1pt]\)
\(\\[10pt]\hspace{0.5em}\) – After n years she owes:
\(\\[7pt]{\scriptsize {u}_{n} = \ 40 000 \times {1.08}^{n} \ – P(1 + 1.08 + {1.08}^{2} + \cdots + {1.08}^{n-1} ) }\)
\(\\[1pt]\)
\(\\[10pt]\) Therefore, the equation in P,
\(\\[10pt] \hspace{4em} {u}_{25} = 0 \)
\(\\[10pt]{\scriptsize 40 000 \times {1.08}^{25} \ – P(1 + 1.08 + {1.08}^{2} + \cdots + {1.08}^{24} ) \ = \ 0 }\)
\(\\[10pt]{\scriptsize P(1 + 1.08 + {1.08}^{2} + \cdots + {1.08}^{24} ) \ = \ 40 000 \times {1.08}^{25} }\)
\(\\[20pt]{\small \hspace{2em} P \ = \ {\large \frac{40 000 \times {1.08}^{25}}{ (1 + 1.08 + {1.08}^{2} + \cdots + {1.08}^{24} ) }} }\)
\(\\[12pt]\) We’ll evaluate \({\small 1 + 1.08 + {1.08}^{2} + \cdots + {1.08}^{24} }\),
\(\\[7pt]\) This is a geometric series with,
\(\\[10pt] \hspace{1.2em} {\small a \ = \ 1 }\)
\(\\[10pt] \hspace{1.2em} {\small r \ = \ 1.08 }\)
\(\\[10pt] \hspace{1.2em} {\small n \ = \ 25 }\)
\(\\[18pt]{\small {S}_{n} \ = \ {\large\frac{a \ ( {r}^{ n } \ – \ 1)}{ r \ – \ 1 } } }\)
\(\\[15pt]{\small {S}_{25} \ = \ {\large\frac{ 1 \ \times \ ( {1.08}^{ 25 } \ – \ 1 )}{1.08 \ – \ 1 } } }\)
\(\hspace{1.7em}{\small \approx \ 73.1059 }\)
\(\\[1pt]\)
\(\\[20pt]{\small P \ = \ {\large \frac{40 000 \times {1.08}^{25}}{ (1 + 1.08 + {1.08}^{2} + \cdots + {1.08}^{24} ) }} }\)
\(\\[15pt]{\small P \ = \ {\large \frac{40 000 \times {1.08}^{25}}{ 73.1059 }} }\)
\(\\[12pt]\hspace{1.1em}{\small \approx \ 3 \ 747.15 }\)
Hence, the value of each annual payment is $3 747.15

\(\\[1pt]\)
\({\small 10.\enspace}\) The Bank of Utopia offers an interest rate of 100% per annum with various options as to how the interest may be added. Gopal invests $1000 and considers the following options.
\(\\[1pt]\)
\( {\small \hspace{1.2em} \textrm{Option A} \hspace{1em} }\) Interest added annually at the end of the year.
\(\\[1pt]\)
\( {\small \hspace{1.2em} \textrm{Option B} \hspace{1em} }\) Interest of 50% credited at the end of each half-year.
\(\\[1pt]\)
\( {\small \hspace{1.2em} \textrm{Option C, D, E, …} \hspace{1em} }\) The Bank is willing to add interest as often as required, subject to (interest rate) \( {\small \times }\) (number of credits per year) = 100.
\(\\[1pt]\)
Investigate to find the maximum possible amount in Gopal’s account after one year.

Show solution:

The number of compounding affects the maximum possible amount. Obviously, the more the compounding the bigger the possible amount is.
\(\\[1pt]\)
Hence, we will use the continuous compounding formula to calculate the maximum possible amount.
\(\\[1pt]\)
\(\\[12pt]\hspace{3em} A \ = \ P {e}^{ r.t } \)
\(\\[7pt]{\small A \ = \ }\) the amount after compounding
\(\\[7pt]{\small P \ = \ }\) the principal amount
\(\\[7pt]{\small e \ = \ }\) the natural or Euler’s number
\(\\[7pt]{\small r \ = \ }\) the annual interest rate
\(\\[12pt]{\small t \ = \ }\) the annual time length of compounding
\(\\[7pt]\) In this example,
\(\\[7pt]{\small P \ = \ 1000 }\)
\(\\[7pt]{\small r \ = \ 100 \% }\)
\(\\[12pt]{\small t \ = \ 1 }\)
\(\\[7pt]\) Plug in the values to find A,
\(\\[10pt] {\small A \ = \ P {e}^{ r.t } }\)
\(\\[10pt]{\small A \ = \ 1000 \ \times \ {e}^{ (1 \ . \ 1) } }\)
\(\\[12pt]{\small \hspace{1.1em} \approx 2 \ 718.28 }\)
Hence, the maximum possible amount in Gopal’s account after one year is $ 2 718.28

\(\\[1pt]\)

PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1. \hspace{0.6em}(\textrm{i}).\hspace{0.7em}}\) The first and second terms of a geometric progression are p and 2p respectively, where p is a positive constant. The sum of the first n terms is greater than 1000p. Show that \({\small {2}^{n} \ \gt \ 1001}\).

\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) In another case, p and 2p are the first and second terms respectively of an arithmetic progression. The nth term is 336 and the sum of the first n terms is 7224. Write down two equations in n and p and hence find the values of n and p.

\({\small 2. \enspace}\) The first three terms of an arithmetic progression are 4, x and y respectively. The first three terms of a geometric progression are x, y and 18 respectively. It is given that both x and y are positive.
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Find the value of x and the value of y.
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\)Find the fourth term of each progression.

\({\small 3. \enspace}\) In an arithmetic progression the first term is a and the common difference is 3. The nth term is 94 and the sum of the first n terms is 1420. Find n and a.

\({\small 4. \enspace}\) A person wants to buy a pension which will provide her with an income of $10 000 at the end of each of the next n years. Show that, with a steady interest rate of 5% per year, the pension should cost her
\(\\[1pt]\)
\({\small \$10 \ 000 } \big(\frac{1}{1.05} + \frac{1}{{1.05}^{2}} + \frac{1}{{1.05}^{3}} + \cdots + \frac{1}{{1.05}^{n}} \big) . \)
\(\\[1pt]\)
Find a simple formula for calculating this sum, and find its value when n = 10, 20, 30, 40, 50.

\({\small 5. \enspace}\) As part of a fund-raising campaign, I have been given some books of raffle tickets to sell. Each book has the same number of tickets and all the tickets I have been given are numbered in sequence. The number of the ticket on the front of the 5th book is 205 and that on the front of the 19th book is 373.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) By writing the number of the ticket on the front of the first book as a and the number of tickets in each book as d, write down two equations involving a and d.
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}\) From these two equations find how many tickets are in each book and the number on the front of the first book I have been given.
\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.6em}}\) The last ticket I have been given is numbered 492. How many books have I been given?

\({\small 6. \enspace}\) A tank is filled with 20 litres of water. Half the water is removed and replaced with anti-freeze and thoroughly mixed. Half this mixture is then removed and replaced with anti-freeze. This process continues.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) Find the first five terms in the sequence of amounts of water in the tank at each stage.
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the first five terms in the sequence of amounts of anti-freeze in the tank at each stage.
\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}\) Is either of these sequences geometric? Explain.

\({\small 7. \enspace}\) A company producing salt from sea water changed to a new process. The amount of salt obtained each week increased by 2% of the amount obtained in the preceding week. It is given that in the first week after the change the company obtained 8000 kg of salt.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) Find the amount of salt obtained in the 12th week after the change.
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the total amount of salt obtained in the first 12 weeks after the change.

\({\small 8. \enspace}\) The common ratio of a geometric progression is 0.99. Express the sum of the first 100 terms as a percentage of the sum to infinity, giving your answer correct to 2 significant figures.

\({\small 9. \hspace{0.6em}(\textrm{a}).\hspace{0.7em}}\) Each year, the value of a certain rare stamp increases by 5% of its value at the beginning of the year. A collector bought the stamp for $10 000 at the beginning of 2005. Find its value at the beginning of 2015 correct to the nearest $100.

\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) The sum of the first n terms of an arithmetic progression is \({\small \frac{1}{2}n(3n+7)}\). Find the 1st term and the common difference of the progression.

\({\small 10. \hspace{0.4em}(\textrm{a}).\hspace{0.7em}}\) An organ has 8 tubes. The lengths of these tubes are a geometrical progression, and the length of the shortest tube is exactly half that of the longest which is 66 cm long. What is the total length of all the tubes in the organ to the nearest cm?

\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) Determine how long it will take to repay an interest free loan $6 625 by monthly payments initially of $10 and increasing by $5 each month. How much is the final payment?

If the loan is $10 000, find the smallest number of months that would be needed to repay it.

\(\\[1pt]\)

As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .