Geometric Progression Concepts

1. (n^{th}) Term and Finite Sum of a GP

• Formula: For GP (a, ar, ar^2,\dots):

$$a_n=ar^{n-1}$$ $$S_n=\frac{a(r^n-1)}{r-1},\quad r\ne1$$

If (r<1), an equivalent form is (S_n=\frac{a(1-r^n)}{1-r}).

• Example: For the GP (3,6,12,\dots), find the 6th term and the sum of the first 6 terms. Solution:

$$a=3,\quad r=2$$

6th term:

$$a_6=3(2)^5=96$$

Sum:

$$S_6=\frac{3(2^6-1)}{2-1} =3(63)=189$$

Answer: 6th term (=96), sum (=189).

ASCII:

3 → 6 → 12 → 24 → 48 → 96
    ×2   ×2    ×2    ×2    ×2

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2. Infinite GP Series

• Formula: For an infinite GP,

$$S_\infty=\frac{a}{1-r},\qquad |r|<1$$

• Example: Find the sum:

$$1+\frac12+\frac14+\frac18+\cdots$$

Solution:

$$a=1,\quad r=\frac12$$

Since (|r|<1), the infinite sum exists.

$$S_\infty=\frac{1}{1-\frac12}=2$$

Answer: (\boxed{2})

ASCII:

1 + 1/2 + 1/4 + 1/8 + ...
│     │      │      │
a     ar     ar²    ar³

r = 1/2

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3. Finding the Common Ratio or First Term

• Formula: For consecutive GP terms,

$$r=\frac{a_{n+1}}{a_n}$$

and

$$a_n=ar^{n-1}$$

• Example: The 3rd and 6th terms of a GP are 12 and 96 respectively. Find the common ratio. Solution:

$$a_3=ar^2=12$$ $$a_6=ar^5=96$$

Divide:

$$\frac{ar^5}{ar^2}=\frac{96}{12}$$ $$r^3=8$$ $$\boxed{r=2}$$

ASCII:

a₁ → a₂ → a₃ → a₄ → a₅ → a₆
       ×r   ×r   ×r   ×r   ×r

a₃ = 12
a₆ = 96

96/12 = r³ = 8
r = 2

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4. Geometric Mean

• Formula: The geometric mean of two positive numbers (a,b) is

$$GM=\sqrt{ab}$$

If (a,G,b) are consecutive GP terms:

$$G^2=ab$$

• Example: Find the geometric mean between 4 and 25. Solution:

Let the geometric mean be (G).

$$G=\sqrt{4\times25}$$ $$G=\sqrt{100}=10$$

Answer: (\boxed{10})

ASCII:

4  →  10  →  25
     GM

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5. Finding Missing Terms in a GP

• Formula: In a GP, if three consecutive terms are (a,b,c),

$$b^2=ac$$

• Example: Find (x) if (3,x,27) are consecutive terms of a GP. Solution:

$$x^2=3\times27=81$$ $$x=\pm9$$

Therefore, the possible values are:

$$\boxed{x=9\text{ or }-9}$$

For a positive GP, (x=9).

ASCII:

3  →  x  →  27
       ↑
     x² = 3×27

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6. Growth and Decay Models

• Formula: For growth at (r%) per period:

$$P_n=P_0\left(1+\frac r{100}\right)^n$$

For decay/depreciation:

$$P_n=P_0\left(1-\frac r{100}\right)^n$$

• Example: A machine worth ₹50,000 depreciates by 10% per year. Find its value after 2 years. Solution:

$$P_0=50000,\quad r=10$$ $$P_2=50000(1-0.10)^2$$ $$=50000(0.9)^2$$ $$=50000(0.81)=₹40,500$$

Answer: (\boxed{₹40,500})

ASCII:

₹50,000
   ↓ −10%
₹45,000
   ↓ −10%
₹40,500

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7. Recursive GP Formulations

• Formula: Each term is obtained by multiplying the previous term by the common ratio:

$$a_n=ra_{n-1}$$

with

$$a_1=a$$

• Example: A sequence has (a_1=5) and (a_n=3a_{n-1}). Find (a_5). Solution:

$$a_1=5$$ $$a_2=3(5)=15$$ $$a_3=3(15)=45$$ $$a_4=3(45)=135$$ $$a_5=3(135)=405$$

Answer: (\boxed{405})

ASCII:

5 → 15 → 45 → 135 → 405
   ×3    ×3     ×3     ×3

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8. Sum of GP When (r=1)

• Formula: If (r=1), every term is (a), so

$$S_n=na$$

• Example: Find the sum of the first 20 terms of the GP (7,7,7,7,\dots). Solution:

Here,

$$a=7,\quad r=1,\quad n=20$$

Hence,

$$S_{20}=20(7)=140$$

Answer: (\boxed{140})

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Advanced Variants

9. Sum of a GP Using the First and Last Terms

• Formula:

$$S_n=\frac{a(r^n-1)}{r-1}$$

Since (a_n=ar^{n-1}), another useful form is

$$S_n=\frac{a_n r-a}{r-1}$$

• Example: Find the sum of (2+6+18+\cdots+486). Solution:

$$a=2,\quad r=3$$

Find (n):

$$486=2(3^{n-1})$$ $$3^{n-1}=243=3^5$$

So,

$$n=6$$

Therefore,

$$S_6=\frac{2(3^6-1)}{3-1}$$ $$=\frac{2(729-1)}2=728$$

Answer: (\boxed{728})

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10. Infinite GP with Negative Common Ratio

• Formula:

$$S_\infty=\frac{a}{1-r},\qquad |r|<1$$

The formula also works when (r<0).

• Example: Find

$$10-5+2.5-1.25+\cdots$$

Solution:

$$a=10,\quad r=-\frac12$$

Since

$$|r|=\frac12<1$$

the sum exists.

$$S_\infty=\frac{10}{1-(-1/2)} =\frac{10}{3/2} =\frac{20}{3}$$

Answer: (\boxed{\frac{20}{3}})

ASCII:

10 → −5 → 2.5 → −1.25 → ...
     ×(−1/2) each time

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11. Sum of Odd/Even Powers in a GP

• Formula: For (a,ar,ar^2,\dots):

$$\text{Odd-position terms: }a+ar^2+ar^4+\cdots$$ $$\text{Even-position terms: }ar+ar^3+ar^5+\cdots$$

• Example: Find the sum of the first 4 odd-position terms of (2,4,8,16,32,\dots). Solution:

Odd-position terms are:

$$2,8,32,128$$

This itself is a GP with:

$$a=2,\quad r=4,\quad n=4$$ $$S_4=\frac{2(4^4-1)}{4-1}$$ $$=\frac{2(256-1)}3 =\frac{510}{3}=170$$

Answer: (\boxed{170})

ASCII:

Original:  2 → 4 → 8 → 16 → 32 → 64 → 128
Positions: 1   2   3    4    5    6     7

Odd:       2       8         32         128
           └── ×4 ──┘

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12. GP Word Problems: Repeated Percentage Change

• Formula: If a quantity changes by the same percentage (r%) repeatedly, its successive values form a GP:

$$V_n=V_0\left(1\pm\frac r{100}\right)^n$$

• Example: A company’s production increases by 20% every year. If current production is 10,000 units, find production after 3 years. Solution:

$$V_0=10000,\quad r=20$$ $$V_3=10000(1.2)^3$$ $$=10000(1.728)$$ $$=17,280$$

Answer: (\boxed{17,280\text{ units}})

ASCII:

10,000 → 12,000 → 14,400 → 17,280
          ×1.2      ×1.2      ×1.2