Geometric Progression IIT JEE Study Material

A geometric progression, or a geometric sequence, is a sequence of numbers in which the ratio of consecutive terms is always constant. This constant term is called the common ratio. Also, the first term of a geometric progression is always a non-zero number. Let a be the 1st term and r be the common ratio of a GP; then, the general form of a geometric progression can be written as

a, ar, ar², ar³ , ar⁴, …. ar^n-1
(Where, n = Total number of terms in a GP)

Therefore, the nth term of a GP is given by

t_n = ar^n-1

Where, common ratio r = t_n/(t_n – 1)

The sum of the first n terms of a geometric progression is given by

Case 1: When r = 1, S_n = na [a = First term]

Case 2: When r > 1,

The Sum of an Infinite GP

When n → ∞, rⁿ → 0 [ If |r| < 1 ]

Therefore, if |r| < 1, then,

Properties of Common Ratio (r)

The value of the common ratio (r) determines whether the GP is decreasing, increasing, positive or negative.

Case 1: If r > 1, then the GP will show exponential growth towards positive infinity.

Case 2: If r < – 1, the GP will show an unsigned (alternating sign) exponential growth towards infinity.

Case 3: If -1 < common ratio (r) < 1 and r ≠ 0, then the GP will decay exponentially toward zero.

Properties of a Geometric Progression

1. If m, n, and o are any three consecutive terms of a geometric sequence, then n² = mo.

2. If t₁, t₂, t₃, t₄, t₅,….., t_n-3, t_n-2, t_n-1, t_n are in geometric progression, then the product of the terms equidistant from the beginning and the end is constant. Also, it is equal to the product of the 1st and the last term.

3. If t₁, t₂, t₃, t₄,….., t_n are in GP and t₁, t₂, t₃, t₄,….., t_n >0, then log t₁, log t₂, log t₃, log t₄,….., log t_n are in AP and vice versa.

4. If each term of a geometric progression is either divided, multiplied or raised to power by the same non-zero number, then the resulting sequence will also form a GP.

5. If t₁, t₂, t₃, t₄,….., t_n and p₁, p₂, p₃, p₄,….., p_n are two geometric progressions with the common ratio as a and b, respectively, then the resulting sequence t₁p₁, t₂p₂, t₃p₃, t₄p₄,….., t_np_n will also form a geometric sequence with a × b as the common ratio (r).

Similarly,

6. Three terms of a geometric progression can be assumed as a/r, a, and ar. Similarly, four terms of a geometric sequence can be taken as a/r², a/r, a, and ar.

Geometric Progression IIT JEE Problems

Example 1: Find the mth term of a geometric progression if its (m + n)^th term is p and (m – n)^th term is q.

Solution:

Given: T_(m+n) = ar^(m+n-1) = p . . . . . (1)

And, T_(m-n) = ar^(m-n-1)  = q . . . . . (2)

On dividing equations (1) and (2), we will get,

i.e.,

Therefore,

Hence,

Now, using equations (1) and (3), we will get

Therefore,

Now, the mth term = a r^(m-1)

Therefore, the mth term of a geometric progression is √(pq). 

Example 2: Find the common ratio (r) of a geometric progression if all its terms are positive and each term is equal to the sum of its consecutive next two terms.

Solution:

Let a, ar, ar² be the terms of GP.

According to the given condition, a = ar +ar²

⇒ r² + r – 1 = 0

Now, by using the quadratic formula,

Therefore, the common ratio (r) = (-1 + √5)/2 (Since each term of GP is positive.)

Example 3: Find the product of the first five terms of a GP, if its third term is 4.

Solution:

Let a be the first term of GP and r be the common ratio.

Given third term = 4

⇒ ar² = 4

Product of first 5 terms = a(ar)(ar²)(ar³)(ar⁴)

= a⁵r¹⁰

= (ar²)⁵

= 4⁵

= 1024

Therefore, the product of the first five terms of a GP is 1024.

Example 4: Find the sum of the series 5 + 55 + 555 + . . . . . n terms.

Solution:

5 + 55 + 555 + . . . . . . + n = (5/9)( 9 + 99 + 999 +  . . . . . n terms)

= (5/9)(10 – 1 + 10² – 1 + 10³-1+….10ⁿ-1)

= (5/9)(10 + 10² + 10³ + ..10ⁿ – (1 + 1 +1+ … n terms)

Therefore, 

Example 5: Find the value of m if, (10)⁹ + 2 (11)¹ (10)⁸ + 3 (11)² (10)⁷ + 4 (11)³ (10)⁶ + . . . . + 10(11)⁹ = m (10)⁹.

Solution:

Given: (10)⁹ + 2 (11)¹ (10)⁸ + 3 (11)² (10)⁷ + 4 (11)³ (10)⁶ + . . . + 10(11)⁹ = m (10)⁹ . . . . . (1)

On dividing equation (1) by 10⁹, we will get

Now, multiplying the above equation by 11/10, we get

Now, Equation (3) – Equation (2), we will get

Since, a = 1, n = 10 and r = 11/10

Or,

Or,

Therefore, the value of m = 100

Geometric Progression Videos

Problem Based on Sum of Infinite Terms of a GP