Introduction to Geometric Progression (GP)

A Geometric Progression (GP) is a sequence of non-zero numbers in which the ratio of any term to its preceding term is always constant. This constant is called the common ratio (r).

  • Example: The sequence 3, 6, 12, 24, … is a GP with a first term a=3a=3 and a common ratio r=2r=2.

Key Formulas for GP:

  • General Term (n-th term): The n-th term of a GP is given by:

an=arn−1a_n = ar^{n-1}

where 'a' is the first term and 'r' is the common ratio.
  • Sum of first n terms: The sum of the first n terms of a GP is given by:

Sn=a(rn−1)r−1orSn=a(1−rn)1−r(for r≠1)S_n = \frac{a(r^n-1)}{r-1} \quad \text{or} \quad S_n = \frac{a(1-r^n)}{1-r} \quad (\text{for } r \ne 1)

  • Sum of an Infinite GP: The sum of an infinite GP exists (converges) only if the absolute value of the common ratio is less than 1 (i.e., ∣r∣<1|r| < 1). The sum is given by:

S∞=a1−rS_{\infty} = \frac{a}{1-r}

Properties of GP:

  • If each term of a GP is multiplied or divided by the same non-zero constant, the resulting sequence is also a GP.

  • The product of terms equidistant from the beginning and end is constant.

  • If we take the logarithm of each term of a GP (with positive terms), the resulting sequence is an AP.

  • Selection of Terms: For problems involving products, it is convenient to select terms as:

    • 3 terms: a/r,a,ara/r, a, ar
    • 4 terms: a/r3,a/r,ar,ar3a/r^3, a/r, ar, ar^3
    • 5 terms: a/r2,a/r,a,ar,ar2a/r^2, a/r, a, ar, ar^2

  • Geometric Mean (GM): The GM of two positive numbers a and b is ab\sqrt{ab}. Inserting 'n' geometric means between a and b means creating a GP of n+2 terms.

Example 1: Finding a Specific Term

Question: Find the 7th term of the GP: 2, 6, 18, …

Solution: Here, the first term a=2a=2 and the common ratio r=6/2=3r = 6/2=3. We need to find the 7th term, so n=7. Using the formula an=arn−1a_n = ar^{n-1}:

a7=2⋅(3)7−1=2⋅36=2⋅729=1458a_7 = 2 \cdot (3)^{7-1} = 2 \cdot 3^6 = 2 \cdot 729 = 1458

Example 2: Sum of a Finite GP

Question: How many terms of the GP 3, 3/2, 3/4, … are needed to give the sum 3069/512?

Solution: Here, a=3a=3 and r=(3/2)/3=1/2r = (3/2)/3 = 1/2. Let the sum of n terms be Sn=3069/512S_n = 3069/512. Using the sum formula Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r}:

3069512=3(1−(1/2)n)1−1/2=3(1−(1/2)n)1/2=6(1−1/2n)\frac{3069}{512} = \frac{3(1 - (1/2)^n)}{1-1/2} = \frac{3(1-(1/2)^n)}{1/2} = 6(1-1/2^n)

30693072=1−12n  ⟹  10231024=1−12n\frac{3069}{3072} = 1 - \frac{1}{2^n} \implies \frac{1023}{1024} = 1 - \frac{1}{2^n}

12n=1−10231024=11024\frac{1}{2^n} = 1 - \frac{1023}{1024} = \frac{1}{1024}

Since 1024=2101024=2^{10}, we have n=10n=10. 10 terms are needed.

Example 3: Sum of an Infinite GP

Question: Find the sum of the infinite series 1−1/3+1/9−1/27+...1 - 1/3 + 1/9 - 1/27 + ...

Solution: This is an infinite GP with first term a=1a=1 and common ratio r=−1/3r=-1/3. Since ∣r∣=∣−1/3∣=1/3<1|r|=|-1/3|=1/3 < 1, the sum exists. Using the formula S∞=a1−rS_{\infty} = \frac{a}{1-r}:

S∞=11−(−1/3)=11+1/3=14/3=34S_{\infty} = \frac{1}{1 - (-1/3)} = \frac{1}{1+1/3} = \frac{1}{4/3} = \frac{3}{4}

Example 4: Selection of Terms

Question: The sum of three numbers in a GP is 39/10, and their product is 1. Find the numbers.

Solution: Let the numbers be a/r,a,ara/r, a, ar.

Product: (a/r)(a)(ar)=1  ⟹  a3=1  ⟹  a=1(a/r)(a)(ar) = 1 \implies a^3 = 1 \implies a = 1.

Sum: 1r+1+r=3910\frac{1}{r} + 1 + r = \frac{39}{10}.

1+r+r2r=3910  ⟹  10r2+10r+10=39r\frac{1+r+r^2}{r} = \frac{39}{10} \implies 10r^2 + 10r + 10 = 39r

10r2−29r+10=0  ⟹  (5r−2)(2r−5)=010r^2 - 29r + 10 = 0 \implies (5r-2)(2r-5) = 0

So, r=2/5r=2/5 or r=5/2r=5/2.

If r=2/5r=2/5, the numbers are 5/2, 1, 2/5. If r=5/2r=5/2, the numbers are 2/5, 1, 5/2. The numbers are 2/5, 1, 5/2.