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Quantitative Aptitude: Progressions

1. Concept Definitions & Explanations

  • Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. E.g., $3, 7, 11, 15, \dots$ (common difference $d = 4$).
  • Geometric Progression (GP): A sequence where the ratio of consecutive terms is constant. E.g., $2, 6, 18, 54, \dots$ (common ratio $r = 3$).
  • Harmonic Progression (HP): A sequence of numbers whose reciprocals form an Arithmetic Progression. E.g., $\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \dots$ (reciprocals $2, 5, 8, 11, \dots$ form an AP).

2. Key Formulas & Shortcuts

Arithmetic Progression (AP)

Let $a = \text{first term}$, $d = \text{common difference}$.

  • $n^{\text{th}}$ term ($t_n$): $$t_n = a + (n-1)d$$
  • Sum of first $n$ terms ($S_n$): $$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)$$ (where $l$ is the last term)
  • Arithmetic Mean (AM) between two numbers $a$ and $b$: $$\text{AM} = \frac{a + b}{2}$$

Geometric Progression (GP)

Let $a = \text{first term}$, $r = \text{common ratio}$.

  • $n^{\text{th}}$ term ($t_n$): $$t_n = a \cdot r^{n-1}$$
  • Sum of first $n$ terms ($S_n$): $$S_n = \frac{a(r^n - 1)}{r - 1} \text{ (if } r > 1\text{), or } S_n = \frac{a(1 - r^n)}{1 - r} \text{ (if } r < 1\text{)}$$
  • Sum of an Infinite GP ($S_{\infty}$) (for $-1 < r < 1$): $$S_{\infty} = \frac{a}{1 - r}$$
  • Geometric Mean (GM) between two numbers $a$ and $b$: $$\text{GM} = \sqrt{a \cdot b}$$

Harmonic Progression (HP)

  • $n^{\text{th}}$ term ($t_n$): $$t_n = \frac{1}{a + (n-1)d}$$ (where $a$ and $d$ belong to the corresponding reciprocal AP)
  • Harmonic Mean (HM) between two numbers $a$ and $b$: $$\text{HM} = \frac{2ab}{a+b}$$

Relation between Means (AM, GM, HM)

  • Mathematical Relation: $$\text{GM}^2 = \text{AM} \times \text{HM}$$
  • Inequality: $$\text{AM} \ge \text{GM} \ge \text{HM}$$ (equality holds only if all numbers in the set are identical)

3. Practice Problems

Problem 1

Find the $15^{\text{th}}$ term of the Harmonic Progression: $\frac{1}{3}, \frac{1}{7}, \frac{1}{11}, \frac{1}{15}, \dots$

Problem 2

Find the sum of the infinite geometric series: $9 - 3 + 1 - \frac{1}{3} + \dots$

Problem 3

The arithmetic mean of two numbers is 10 and their geometric mean is 8. Find the numbers.

Problem 4

How many terms are there in the Arithmetic Progression: $20, 25, 30, \dots, 135$?

Problem 5

Find the sum of all natural numbers between 100 and 300 which are exactly divisible by 4.

Problem 6

If the 3rd and 6th terms of a Geometric Progression are 12 and 96 respectively, find the first term and the common ratio.

Problem 7

Insert three arithmetic means between 3 and 19.

Problem 8

Find the sum of the first 10 terms of the geometric series: $2, 6, 18, 54, \dots$

Problem 9

If the sum of the first $n$ terms of an AP is given by $S_n = 3n^2 + 5n$, find the $10^{\text{th}}$ term of this AP.

Problem 10

Find the sum of the infinite geometric progression: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$

4. Step-by-Step Solutions

Solution 1

  1. Identify the reciprocal sequence:
    • The terms of the HP are $\frac{1}{3}, \frac{1}{7}, \frac{1}{11}, \frac{1}{15}, \dots$
    • The reciprocals are $3, 7, 11, 15, \dots$
  2. Verify and analyze the reciprocal AP:
    • First term $a = 3$.
    • Common difference $d = 7 - 3 = 4$.
  3. Find the $15^{\text{th}}$ term of the AP: $$t_{15} = a + (15-1)d = 3 + 14(4) = 3 + 56 = 59$$
  4. Reciprocate to find the HP term: $$\text{HP term } t_{15} = \frac{1}{59}$$
  5. Answer: The $15^{\text{th}}$ term is $\frac{1}{59}$.

Solution 2

  1. Identify parameters of the GP:
    • Series: $9, -3, 1, -\frac{1}{3}, \dots$
    • First term $a = 9$.
    • Common ratio $r = \frac{-3}{9} = -\frac{1}{3}$.
  2. Check for convergence:
    • Since $|r| = |-\frac{1}{3}| = \frac{1}{3} < 1$, the infinite sum converges.
  3. Apply the Infinite GP Sum Formula: $$S_{\infty} = \frac{a}{1 - r} = \frac{9}{1 - (-1/3)} = \frac{9}{1 + 1/3} = \frac{9}{4/3} = 9 \times \frac{3}{4} = \frac{27}{4} = 6.75$$
  4. Answer: The sum of the infinite series is $6.75$ (or $\frac{27}{4}$).

Solution 3

  1. Set up the mean equations:
    • Let the two numbers be $x$ and $y$.
    • $\text{AM} = \frac{x+y}{2} = 10 \implies x + y = 20$.
    • $\text{GM} = \sqrt{xy} = 8 \implies xy = 64$.
  2. Solve the quadratic equation system:
    • From $x+y=20$, we can express $y = 20 - x$.
    • Substitute into $xy = 64$: $$x(20 - x) = 64$$ $$20x - x^2 = 64 \implies x^2 - 20x + 64 = 0$$
  3. Factor the quadratic equation: $$(x - 16)(x - 4) = 0 \implies x = 16 \text{ or } x = 4$$
    • If $x=16$, then $y=4$. If $x=4$, then $y=16$.
  4. Answer: The two numbers are 16 and 4.

Solution 4

  1. Identify the progression variables:
    • First term $a = 20$.
    • Common difference $d = 25 - 20 = 5$.
    • Last term $t_n = 135$.
  2. Apply the $n^{\text{th}}$ term formula:
    • $t_n = a + (n - 1)d$
    • $135 = 20 + (n - 1)5$
    • $115 = (n - 1)5 \implies n - 1 = 23 \implies n = 24$.
  3. Answer: There are 24 terms in the progression.

Solution 5

  1. Find the first and last term in the range:
    • The natural numbers between 100 and 300 divisible by 4 are $104, 108, 112, \dots, 296$.
    • First term $a = 104$.
    • Common difference $d = 4$.
    • Last term $l = 296$.
  2. Find the number of terms ($n$):
    • $l = a + (n - 1)d \implies 296 = 104 + (n - 1)4$
    • $192 = 4(n - 1) \implies n - 1 = 48 \implies n = 49$.
  3. Calculate the sum ($S_n$):
    • $S_n = \frac{n}{2}(a + l) = \frac{49}{2}(104 + 296) = \frac{49}{2}(400) = 49 \times 200 = 9800$.
  4. Answer: The sum of the numbers is 9,800.

Solution 6

  1. Set up the GP term equations:
    • $t_n = a \cdot r^{n-1}$.
    • $t_3 = a \cdot r^2 = 12$.
    • $t_6 = a \cdot r^5 = 96$.
  2. Find common ratio $r$ by dividing the equations:
    • $\frac{a \cdot r^5}{a \cdot r^2} = \frac{96}{12}$
    • $r^3 = 8 \implies r = 2$.
  3. Find the first term $a$:
    • $a \cdot (2)^2 = 12 \implies 4a = 12 \implies a = 3$.
  4. Answer: The first term is 3 and the common ratio is 2.

Solution 7

  1. Understand insertion of AMs:
    • Let the three arithmetic means be $A_1, A_2, A_3$ between 3 and 19.
    • The sequence $3, A_1, A_2, A_3, 19$ forms an Arithmetic Progression.
  2. Find the common difference $d$:
    • Total terms $n = 5$.
    • First term $a = 3$, last term $t_5 = 19$.
    • $t_5 = a + 4d \implies 19 = 3 + 4d \implies 16 = 4d \implies d = 4$.
  3. Calculate the means:
    • $A_1 = a + d = 3 + 4 = 7$
    • $A_2 = a + 2d = 3 + 8 = 11$
    • $A_3 = a + 3d = 3 + 12 = 15$
  4. Answer: The three arithmetic means are 7, 11, and 15.

Solution 8

  1. Identify the GP parameters:
    • First term $a = 2$.
    • Common ratio $r = \frac{6}{2} = 3$.
    • Number of terms $n = 10$.
  2. Apply GP sum formula (since $r > 1$):
    • $S_n = \frac{a(r^n - 1)}{r - 1}$
    • $S_{10} = \frac{2(3^{10} - 1)}{3 - 1} = \frac{2(3^{10} - 1)}{2} = 3^{10} - 1$.
  3. Calculate $3^{10}$:
    • $3^{10} = 59049$.
    • $S_{10} = 59049 - 1 = 59048$.
  4. Answer: The sum of the first 10 terms is 59,048.

Solution 9

  1. Find the relation between sum and terms:
    • The $n^{\text{th}}$ term $t_n$ is given by $t_n = S_n - S_{n-1}$.
  2. Calculate $S_{10}$ and $S_9$:
    • $S_{10} = 3(10)^2 + 5(10) = 300 + 50 = 350$.
    • $S_9 = 3(9)^2 + 5(9) = 243 + 45 = 288$.
  3. Calculate the $10^{\text{th}}$ term:
    • $t_{10} = S_{10} - S_9 = 350 - 288 = 62$.
  4. Answer: The $10^{\text{th}}$ term of the AP is 62.

Solution 10

  1. Identify the parameters of the GP:
    • First term $a = 1$.
    • Common ratio $r = \frac{1}{2}$.
  2. Apply the sum formula for infinite GP:
    • Since $|r| = \frac{1}{2} < 1$, we use: $$S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$$
  3. Answer: The sum of the infinite GP is 2.