^{th}term by the n

^{th}term is the same as that obtained when the n

^{th}term divided by the (n – 1)

^{th}term.

This common value is called the **common ratio(r)**.

For example, take a sequence of numbers as: 2, 4, 8, 16 Now 4/2=8/4=16/8= 2. So this sequence of numbers forms a GP with a common ratio = 2.

## General form of a GP

Consider a GP with first term as ‘a’ and common ratio as ‘r’.

1^{st} term = a

2^{nd} term = ar

3^{rd} term = ar^{2}

4^{th} term = ar^{3}

Proceeding this way,

** n ^{th} term = T_{n}= ar^{n-1}**

### Sum of GP: Formula

The n^{th} term of a geometric progression is given by T_{n}=ar^{n-1},

So sum to ‘n’ terms,

S_{n} = a + ar + ar^{2} + ar^{3} + … + ar^{n-1}

= a (1 + r + r^{2} + r^{3} + … + r^{n-1})

= a ((1- r^{n} ))/((1-r) ) if r < 1 & a ((r^{n-1}))/((r-1) ) if r > 1

The sum of geometric series up to n terms is given by,

S_{n}= a(1- r^{n} )/((1-r) ) (1 > r)

⇒**S _{n }= a ((r^{n-1}))/((r-1) ) (1 < r)**

### Geometric Progression Questions

**Question:** Find the eight term of the series: 1/3 – 1/6 + 1/12 – 1/24

**a)** 1/384

**b)** 1/192

**c)** (-1)/192

**d)** (-1)/384

**e)** (-1)/128

**Solution:**

The series is a GP with common ratio r = (((-1)/6)/(1/3)) = ((-1)/2), Eight term = ar^{7} = (1/3) × ((-1)/2)^{7} = (-1)/384

**Hence option (d)**

**Question:** Find the sum of the first five terms of the series: 3, 12, 48…

**a)** 512

**b)** 1024

**c)** 198

**d)** 343

**e)** 128

Solution:

S_{n}=[a(r^{n}– 1) ]/(r – 1)

In the question a = 3, r = 4 & n = 5

S_{5}= 3 ×(4^{5}– 1)/(4 – 1)= 1024

**Hence Option (b)**

**Question:** The sum of first two terms of a G.P. is 5/3 and the sum to infinite terms is 3. If the GP has a positive common ratio, what is the first term?

**a)** 1

**b)** 6

**c)** 9

**d)** 3/5

**e)** can’t be determined

**Solution:**

The first two terms = a, ar a + ar = 5/3

a/(1 – r)= 3 ⇒ a = 3(1-r)a(1 + r)=5/3

⇒3(1-r)(1 + r)=5/3⇒1 – r^{2} = 5/9 ?r^{2} = 4/9 ? r = 2/3

∴ a = 3(1-2/3)= 1

So, first term is 1.

**Hence option (a)**

**Question:** The sum of an infinite GP whose common ratio is positive and numerically less than 1 is 32 and the sum of first two terms is 24.

What will be the third term?

**Solution:**

S = a/(1 – r) for an infinite GP with r < 1

⇒32 = a/(1 – r) ? a = 32 (1 – r)

First term = a

Second term = ar

a + ar = 24 ? a = 24/(1 + r)

⇒32(1 – r) = 24/( 1 + r)

⇒1 – r^{2} = 24/32 = 3/4

⇒r^{2} = 1 – 3/4 = 1/4 ? r = 1/2

So a = 32(1/2) = 16

Third term = ar2 = (16)(1/4)= 4.

**Question:** Determine the first term of a GP, the sum of whose first term and third term is 40 and the sum of whose second term and fourth term is 80.

**Solution:**

Suppose the GP has first term as ‘a’ and common ratio as ‘r’

First term = a

Second term = ar

Third term = ar^{2}

Fourth term = ar^{3}

a + ar^{2} = 40 ? a(1 + r^{2}) = 40 —(i)

ar + ar^{3} = 80 ? ar ( 1 + r^{2}) = 80 —(ii)

dividing (ii) by (i)

r = 2

a = 40/((1+4) ) = 8

So, the first term = 8.