Progressions — Concept Explanation

A progression is an ordered sequence of numbers following a definite pattern or rule. The most common progressions are:

  • Natural-number series (simple summation patterns)
  • Arithmetic Progression (A.P.) – constant difference
  • Geometric Progression (G.P.) – constant ratio
Each has its own formula for the nth term and for the sum of terms.

I. Sum of Natural Numbers, Squares, and Cubes

These fundamental results are direct consequences of arithmetic progressions and are often used in algebraic manipulation and series evaluation.

1. Sum of first n natural numbers:

\[ 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} \]

2. Sum of squares of first n natural numbers:

\[ 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6} \]

3. Sum of cubes of first n natural numbers:

\[ 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \]

Example 1:

Find the sum of squares of the first 15 positive integers.

\[ S = \frac{15(16)(31)}{6} = 1240. \]

Example 2:

Find the sum of squares of the second 15 integers, i.e., \(16^2 + 17^2 + \dots + 30^2.\)

Use identity: \[ \sum_{k=m+1}^{n+m} k^2 = \sum_{k=1}^{n+m}k^2 – \sum_{k=1}^{m}k^2. \] Hence, \[ S = \frac{30(31)(61)}{6} – \frac{15(16)(31)}{6} = 6785. \]

II. Arithmetic Progression (A.P.)

An Arithmetic Progression is a sequence in which the difference between any two consecutive terms is constant. That difference is called the common difference \(d\).

Formulas:

  • General or nth term: \[ a_n = a + (n – 1)d \]
  • Number of terms: \[ n = \frac{l – a}{d} + 1 \]
  • Sum of first n terms: \[ S_n = \frac{n}{2}[2a + (n – 1)d] = \frac{n(a + l)}{2} \] where \(a\) is the first term and \(l\) is the last term.

Example 3:

The first term of an A.P. is \(-5\), and the common difference \(d=6\). Find the 104th term.

\[ a_{104} = -5 + (104 – 1)\times6 = -5 + 618 = 613. \] Answer: 613.

Example 4:

Find the sum of even integers between 200 and 400 inclusive.

Even integers form an A.P. \(200, 202, 204, \dots, 400\) with \(a=200, d=2, l=400.\)

\[ n = \frac{400-200}{2} + 1 = 101,\quad S_n = \frac{101(200+400)}{2} = 30{,}300. \]

Example 5:

Find the sum of multiples of 7 from 84 to 140 inclusive.

Multiples: \(84,91,98,105,112,119,126,133,140.\) \(a=84,\ l=140,\ d=7,\ n = \frac{140-84}{7}+1 = 9.\)

\[ S_n = \frac{9(84+140)}{2} = 1008. \]

Example 6:

Sequence \(S_1=6, S_2=12, S_n=S_{n-1}+6.\) Find sum of terms from \(S_{13}\) to \(S_{28}\).

\(a_{13} = 6 + (13-1)\times6 = 78,\ a_{28} = 6 + (28-1)\times6 = 168.\) There are \(28-13+1=16\) terms.

\[ S = \frac{16(78+168)}{2} = 1{,}968. \]

Example 7:

If integer \(k\) is the sum of all even multiples of 15 between 295 and 615, find the greatest prime factor of \(k.\)

Even multiples of 15 between 295 and 615 are \(300, 330, 360, 390, 420, 450, 480, 510, 540, 570, 600.\) \(a=300,\ l=600,\ d=30,\ n=11.\)

\[ S = \frac{11(300+600)}{2}=4{,}950. \] Prime-factorize \(4{,}950 = 2 \times 3^2 \times 5^2 \times 11.\) Greatest prime factor = 11.

Example 8:

Five consecutive even integers → \(x-4, x-2, x, x+2, x+4.\) Given: sum of 2nd, 3rd, and 4th = 132.

\[ (x-2) + x + (x+2) = 3x = 132 \Rightarrow x=44. \] Sum of first and last = \((x-4)+(x+4)=2x=88.\)

Answer: 88.

III. Geometric Progression (G.P.)

A Geometric Progression is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed constant ratio \(r\).

Formulas:

  • General term: \[ a_n = a r^{\,n-1} \]
  • Sum of first n terms (for \(r \ne 1\)): \[ S_n = a\,\frac{r^n – 1}{r – 1} \]
  • If the last term \(l = ar^{n-1}\) is known: \[ S_n = \frac{l r – a}{r – 1} \]
  • For an infinite G.P. with \(|r|<1:\) \[ S_\infty = \frac{a}{1 - r}. \]

Example 9:

Sequence: \( \tfrac{1}{8}, \tfrac{1}{4}, \tfrac{1}{2}, \dots \)

Common ratio \(r = 2.\) \(a_{13} = \frac{1}{8}\times2^{12} = 2^9 = 512.\)

Example 10:

First two terms \(640, 160.\) Common ratio \(r = \frac{160}{640} = \frac{1}{4}.\) Sixth term:

\[ a_6 = 640\left(\frac{1}{4}\right)^{5} = 640 \times \frac{1}{1024} = \frac{5}{8}. \] (or 0.625 numerically)

Example 11:

Evaluate \(2 + 2 + 2^2 + 2^3 + \dots + 2^8.\)

\(a=2,\ r=2,\ n=9.\)

\[ S = 2\,\frac{2^9 – 1}{2 – 1} = 2(512 – 1) = 1{,}022. \]

Example 12 (Recursive G.P.):

Series defined by \(S_n = k S_{n-1},\) where \(k\) is constant. Given \(S_1 = 64,\ S_{25} = 192.\)

\[ S_{25} = S_1 k^{24} \Rightarrow 192 = 64 k^{24} \Rightarrow k^{24} = 3 \Rightarrow k = 3^{1/24}. \] \(S_9 = S_1 k^{8} = 64 \times 3^{8/24} = 64 \times 3^{1/3} = 64\sqrt[3]{3}.\)

Answer: \(64\sqrt[3]{3}.\)

IV. Conceptual Summary

  • Arithmetic Progression → constant difference \(d\)
  • Geometric Progression → constant ratio \(r\)
  • Natural-number sums are direct special cases of A.P. formulas
  • For A.P.: \(a_n = a+(n-1)d,\ S_n = \frac{n}{2}(2a+(n-1)d)\)
  • For G.P.: \(a_n = ar^{n-1},\ S_n = a\frac{r^n – 1}{r – 1}\)
  • Infinite G.P. (|r|<1): \(S_\infty = \frac{a}{1-r}\)
  • Know how to reverse: find \(n, a, d, r\) from given \(l, S_n\) or ratio relationships.

Source reference: Based on “GMAT Algebra Topics 4” compiled by Manoj K. Singh (Progression section). Expanded with full reasoning, derivations, and solved illustrations for conceptual clarity.