Geometric Progression (GP)
1 / 41. Identifying the Common Ratio ($r$)
In a GP, we multiply by the same number ($r$) to get the next term.
3, 6, 12, 24, ...
Find the parameters:
First Term ($a$):
Common Ratio ($r$):
Hint: $r = \frac{\text{2nd Term}}{\text{1st Term}}$Watch out! $r$ can be a fraction or negative.
80, -40, 20, -10, ...
$r = \frac{-40}{80} = $
Correct! Always check signs.
2. Finding the $n$-th Term
Formula: $$ u_n = ar^{n-1} $$
Find the 6th term of: $2, 6, 18, \dots$
Here $a=2$, $r=3$, $n=6$.
Step 1: The Power
$$ u_{6} = 2 \times 3^{\text{power}} $$
The power is $(n-1) = $
Step 2: Calculate
$$ u_{6} = 2 \times 3^5 $$
$$ u_{6} = 2 \times 243 $$
$$ u_{6} = \dots $$
Correct. Geometric sequences grow very fast!
3. Sum to Infinity ($S_\infty$)
If $|r| < 1$, the sequence gets smaller and smaller. We can add all terms up to infinity.
Formula: $$ S_\infty = \frac{a}{1-r} $$
Task: Find sum to infinity of $100, 20, 4, \dots$
Step 1: Check r
$r = \frac{20}{100} = $
Step 2: Apply Formula
Is $|0.2| < 1$? Yes. Proceed.
$$ S_\infty = \frac{100}{1 - 0.2} = \frac{100}{0.8} $$
Final Answer:
Perfect! The sum converges to 125.
4. Exam Style: Working Backwards
Problem: A geometric progression has a first term of 40 and a Sum to Infinity of 100. Find the common ratio $r$ and the 2nd term.
Step 1: Set up Equation
$$ S_\infty = \frac{a}{1-r} = 100 $$
$$ \frac{40}{1-r} = 100 $$
Step 2: Rearrange
Multiply by $(1-r)$:
$$ 40 = 100(1-r) $$
Divide by 100:
$$ 0.4 = 1 - r $$
Step 3: Solve for r
$$ r = 1 - 0.4 $$
$$ r = \mathbf{0.6} $$
Step 4: Find 2nd Term
$$ u_2 = a \times r $$
$$ u_2 = 40 \times 0.6 $$
$$ u_2 = \mathbf{24} $$