IIT JEE Main Maths -Unit 6- Arithmetic and Geometric progressions- Study Notes-New Syllabus

IIT JEE Main Maths -Unit 6- Arithmetic and Geometric progressions – Study Notes – New syllabus

IIT JEE Main Maths -Unit 6- Arithmetic and Geometric progressions – Study Notes -IIT JEE Main Maths – per latest Syllabus.

Key Concepts:

Arithmetic Progressions (A.P.)
Geometric Progressions (G.P.)

IIT JEE Main Maths -Study Notes – All Topics

Arithmetic Progressions (A.P.)

An Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is constant.

This constant difference is called the common difference $ d $.

$ a,\, a + d,\, a + 2d,\, a + 3d,\, \dots $

where $ a $ = first term, $ d $ = common difference.

nth Term of an A.P.

The $ n $th term (general term) of an arithmetic progression is given by:

$ a_n = a + (n – 1)d $

where:

$ a $ = first term
$ d $ = common difference
$ n $ = number of terms

Sum of n Terms (S_n)

The sum of the first $ n $ terms of an A.P. is given by:

$ S_n = \dfrac{n}{2}[2a + (n – 1)d] $

or equivalently,

$ S_n = \dfrac{n}{2}(a + l) $

where $ l = a + (n – 1)d $ is the last term.

Arithmetic Mean (A.M.)

If three numbers $ a, A, b $ are in A.P., then:

$ A = \dfrac{a + b}{2} $

The term $ A $ is called the Arithmetic Mean between $ a $ and $ b $.

Properties of A.P.

The difference between any two consecutive terms is constant: $ a_{n+1} – a_n = d $.
If each term of an A.P. is increased, decreased, multiplied, or divided by a constant, the new sequence is also an A.P.
The nth term forms a linear function of $ n $.

Inserting Arithmetic Means

If $ m $ arithmetic means are inserted between $ a $ and $ b $, the sequence becomes an A.P. with:

$ d = \dfrac{b – a}{m + 1} $

and the means are:

$ a + d,\, a + 2d,\, a + 3d,\, \dots,\, a + md $

Important Shortcuts for JEE

If $ a_p, a_q, a_r $ are terms in an A.P., then:
$ a_q – a_p = (q – p)d $
If $ a_p, a_q, a_r $ are in A.P., then $ 2a_q = a_p + a_r $.
Sum of first $ n $ natural numbers: $ 1 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2} $.
Sum of first $ n $ odd numbers: $ 1 + 3 + 5 + \dots = n^2 $.
Sum of first $ n $ even numbers: $ 2 + 4 + 6 + \dots = n(n + 1) $.

Key Takeaways for JEE:

Use $ a_n = a + (n – 1)d $ for the general term.
Use $ S_n = \dfrac{n}{2}[2a + (n – 1)d] $ for total sum.
Check whether $ n $ is an integer when finding number of terms.
Apply A.P. logic in coordinate geometry, progressions, and sequences problems.
For A.P. insertion: $ d = \dfrac{b – a}{m + 1} $.

Example

Find the 10th term of the A.P. $ 5, 8, 11, 14, \dots $

▶️ Answer / Explanation

Given: $ a = 5, d = 3 $

Formula: $ a_n = a + (n – 1)d $

$ a_{10} = 5 + (10 – 1)(3) = 5 + 27 = 32 $

Answer: 32

Example

Find the sum of the first 15 terms of the A.P. $ 2, 5, 8, 11, \dots $

▶️ Answer / Explanation

$ a = 2, d = 3, n = 15 $

$ S_n = \dfrac{n}{2}[2a + (n – 1)d] $

$ S_{15} = \dfrac{15}{2}[4 + 14(3)] = \dfrac{15}{2}[46] = 345 $

Answer: 345

Example

The 3rd term of an A.P. is 7 and the 7th term is 19. Find the first term, common difference, and 20th term.

▶️ Answer / Explanation

$ a_3 = a + 2d = 7 $

$ a_7 = a + 6d = 19 $

Subtracting: $ (a + 6d) – (a + 2d) = 19 – 7 \Rightarrow 4d = 12 \Rightarrow d = 3 $

Substitute in $ a + 2d = 7 \Rightarrow a + 6 = 7 \Rightarrow a = 1 $

$ a_{20} = a + 19d = 1 + 19(3) = 58 $

Answer: $ a = 1,\, d = 3,\, a_{20} = 58 $

Geometric Progressions (G.P.)

A Geometric Progression (G.P.) is a sequence of numbers in which each term (after the first) is obtained by multiplying the preceding term by a fixed number called the common ratio $ r $.

$ a,\, ar,\, ar^2,\, ar^3,\, ar^4,\, \dots $

where $ a $ = first term, $ r $ = common ratio.

nth Term of a G.P.

The $ n $th term of a geometric progression is given by:

$ a_n = ar^{n – 1} $

If $ |r| > 1 $: terms increase (diverging G.P.)
If $ |r| < 1 $: terms decrease (converging G.P.)

Sum of n Terms of a G.P.

Sum of the first $ n $ terms $ S_n $ of a G.P. is given by:

$ S_n = \begin{cases} \dfrac{a(r^n – 1)}{r – 1}, & r \ne 1 \$6pt] na, & r = 1 \end{cases} $

Alternate Form: $ S_n = \dfrac{a(1 – r^n)}{1 – r}, \text{ if } |r| < 1 $

Sum to Infinity (for |r| < 1)

If $ |r| < 1 $, the G.P. converges, and its infinite sum is:

$ S_\infty = \dfrac{a}{1 – r} $

Geometric Mean (G.M.)

If three numbers $ a, G, b $ are in G.P., then:

$ G^2 = ab \quad \Rightarrow \quad G = \sqrt{ab} $

Here $ G $ is the Geometric Mean between $ a $ and $ b $.

Inserting Geometric Means

If $ n $ geometric means are inserted between $ a $ and $ b $, the common ratio is:

$ r = \left( \dfrac{b}{a} \right)^{\dfrac{1}{n + 1}} $

The means will be $ ar, ar^2, ar^3, \dots, ar^n $.

Relation Between A.M., G.M., and H.M.

For any two positive numbers $ a $ and $ b $:

$ A.M. \ge G.M. \ge H.M. $

and $ A.M. \times H.M. = (G.M.)^2 $.

Example

Find the 8th term of the G.P. $ 3, 6, 12, 24, \dots $

▶️ Answer / Explanation

$ a = 3, r = 2 $

Formula: $ a_n = ar^{n – 1} $

$ a_8 = 3(2)^{7} = 3(128) = 384 $

Answer: 384

Example

Find the sum of first 10 terms of the G.P. $ 5, 10, 20, 40, \dots $

▶️ Answer / Explanation

$ a = 5, r = 2, n = 10 $

$ S_n = \dfrac{a(r^n – 1)}{r – 1} = \dfrac{5(2^{10} – 1)}{1} = 5(1023) = 5115 $

Answer: 5115

Example

Find the sum to infinity of the series $ 7 + 4.2 + 2.52 + \dots $

▶️ Answer / Explanation

Here $ a = 7, r = \dfrac{4.2}{7} = 0.6 $

Since $ |r| < 1 $, $ S_\infty = \dfrac{a}{1 – r} = \dfrac{7}{1 – 0.6} = \dfrac{7}{0.4} = 17.5 $

Answer: $ S_\infty = 17.5 $

Important Relations for JEE

If $ a, b, c $ are in G.P. → $ b^2 = ac $
If $ a, b, c, d $ are in G.P. → $ ad = bc $
Product of $ n $ terms of a G.P. = $ (a_1 a_2 a_3 \dots a_n) = (a_1 a_n)^{\frac{n}{2}} $
Sum of infinite G.P. (for $ |r| < 1 $): $ S_\infty = \dfrac{a}{1 – r} $
G.P. → exponential growth or decay — useful in JEE physics (radioactive decay, population, etc.)

Key Takeaways for JEE:

Use $ a_n = ar^{n-1} $ for the nth term.
Use $ S_n = \dfrac{a(r^n – 1)}{r – 1} $ for finite sum.
For infinite G.P. → $ S_\infty = \dfrac{a}{1 – r} $ (only if $ |r| < 1 $).
Product of $ n $ terms = $ (a_1 a_n)^{n/2} $.
Common ratio $ r = \dfrac{a_{n+1}}{a_n} $.
G.P. is linked to exponential functions in calculus and logarithms.

Example

Insert two geometric means between 3 and 24.

▶️ Answer / Explanation

If $ 3, G_1, G_2, 24 $ are in G.P., $ r = \left( \dfrac{24}{3} \right)^{1/3} = 8^{1/3} = 2 $

So the means are $ 3r = 6 $ and $ 3r^2 = 12 $.

Answer: $ 6 $ and $ 12 $

Notes and Study Materials

Examples and Exercise

IIT JEE (Main) Mathematics ,”Sequence & Series” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian

About this unit

Arithmetic and Geometric progressions, insertion of arithmetic.geometric means between two given numbers.The relation between A.M. and G.M.Sum upto n terms of special series: Sn, Sn2, Sn3 Arithmetico – Geometric progression.

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IIT JEE Main Maths -Unit 6- Arithmetic and Geometric progressions- Study Notes-New Syllabus