Solving Problems Involving APs and GPs (HSC SSCE Mathematics Advanced): Revision Notes

Illustrative Example: Identifying an AP

The numbers $10$, $25$, $40$ form an AP because:

• $25 - 10 = 15$
• $40 - 25 = 15$

The differences are equal, so this is an AP.

Illustrative Example: Identifying a GP

The numbers $10$, $20$, $40$ form a GP because:

• $\frac{20}{10} = 2$
• $\frac{40}{20} = 2$

The ratios are equal, so this is a GP.

Worked Example: Finding the Middle Term

Problem: Find the value of $x$ if:

• Part a: $3$, $x$, $12$ form an AP
• Part b: $3$, $x$, $12$ form a GP

Solution for part a (AP):

Since $3$, $x$, $12$ form an AP, we use the condition that consecutive differences are equal:

$x - 3 = 12 - x$

$2x = 15$

$x = 7\frac{1}{2}$

Solution for part b (GP):

Since $3$, $x$, $12$ form a GP, we use the condition that consecutive ratios are equal:

$\frac{x}{3} = \frac{12}{x}$

$x^2 = 36$

$x = 6 \text{ or } x = -6$

Note that for the GP, we get two possible values for $x$. Both are valid because geometric progressions can have negative terms while maintaining a constant ratio.

Worked Example: Finding Terms Using Subtraction

Problem: The third term of an AP is $16$ and the twelfth term is $79$. Find the 41st term.

Solution:

Let the first term be $a$ and the common difference be $d$.

Since $T_3 = 16$:

$a + 2d = 16 \quad \text{(equation 1)}$

Since $T_{12} = 79$:

$a + 11d = 79 \quad \text{(equation 2)}$

Subtracting equation (1) from equation (2):

$9d = 63$

This is the key step that eliminates $a$.

$d = 7$

Substituting $d = 7$ into equation (1):

$a + 14 = 16$

$a = 2$

Now we can find $T_{41}$:

$T_{41} = a + 40d$

$T_{41} = 2 + 40(7)$

$T_{41} = 2 + 280$

$T_{41} = 282$

Worked Example: Finding Terms Using Division

Problem: Find the first term $a$ and the common ratio $r$ of a GP in which the fourth term is $6$ and the seventh term is $162$.

Solution:

Since $T_4 = 6$:

$ar^3 = 6 \quad \text{(equation 1)}$

Since $T_7 = 162$:

$ar^6 = 162 \quad \text{(equation 2)}$

Dividing equation (2) by equation (1):

$\frac{ar^6}{ar^3} = \frac{162}{6}$

$r^3 = 27$

This is the key step that eliminates $a$.

$r = 3$

Substituting $r = 3$ into equation (1):

$a \times 27 = 6$

$a = \frac{6}{27} = \frac{2}{9}$

Therefore, the first term is $\frac{2}{9}$ and the common ratio is $3$.

Worked Example: Two Methods for the Same Problem

Problem: For the geometric sequence $2$, $6$, $18$, $\ldots$

• Part a: Find a formula for the $n$th term
• Part b: Use trial-and-error to find the first term greater than $1000000$
• Part c: Use logarithms to find the first term greater than $1000000$

Solution for part a:

This is a GP with first term $a = 2$ and common ratio $r = 3$.

Using the formula $T_n = ar^{n-1}$:

$T_n = 2 \times 3^{n-1}$

Solution for part b (trial-and-error):

We need to find when $T_n > 1000000$.

Using a calculator to evaluate successive terms:

$T_{12} = 2 \times 3^{11} = 354294$

$T_{13} = 2 \times 3^{12} = 1062882$

Since $T_{12} = 354294 < 1000000$ and $T_{13} = 1062882 > 1000000$, the first term greater than $1000000$ is $T_{13} = 1062882$.

Solution for part c (logarithms):

We need to solve $T_n > 1000000$:

$2 \times 3^{n-1} > 1000000$

$3^{n-1} > 500000$

Taking logarithms of both sides (remembering that if $2^3 = 8$, then $3 = \log_2 8$):

$n - 1 > \log_3 500000$

Using the change-of-base formula:

$n - 1 > \frac{\log_{10} 500000}{\log_{10} 3}$

$n - 1 > 11.94\ldots$

$n > 12.94\ldots$

Since $n$ must be a whole number (as it represents a term position), we need $n \geq 13$.

Therefore, the first term greater than $1000000$ is $T_{13} = 1062882$.

Both methods give the same answer, confirming our result. The trial-and-error method was quick in this case because we only needed to test two values. The logarithm method provides a systematic approach that works regardless of how large the target value is.