# WORD PROBLEMS IN GEOMETRIC SEQUENCE

Word Problems in Geometric Sequence :

Here we are going to see some practice questions of finding geometric progression with given information.

## Word Problems in Geometric Sequence - Questions

Question 1 :

A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?

Solution :

Starting salary   =  60,000

Every year 5% of annual salary is increasing.

Second year salary  =  60000 + 5% of 60000

=  60000(1 + 5%)

Third year salary  =  60000 + 5% of 60000(1 + 5%)

=  60000(1 + 5%)(1 + 5%)

=  60000(1 + 5%)2

By continuing in this way,  salary after 5 years is

=  60000(1 + 5%)5

=  60000(105/100)5

=  60000(1.05)

=  76577

Question 2 :

Sivamani is attending an interview for a job and the company gave two offers to him.

Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.

Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

What is his salary in the 4th year with respect to the offers A and B?

Solution :

Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.

a = 20000, r = 0.06 and n = 5

Starting salary   =  20,000

Every year 6% of annual salary is increasing.

Starting salary  =  20000

Second year salary  =  20000 + 6% of 20000

=  20000(1 + 6%)

By continuing in this way, we get

4th year salary  =  20000(1 + 6%)3

=  20000(1.06)3

=  22820

Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

a = 22000, r = 0.03 and n = 5

Starting salary   =  22,000

Every year 3% of annual salary is increasing.

Starting salary  =  22000

Second year salary  =  22000 + 3% of 22000

=  22000(1 + 3%)

By continuing in this way, we get

4th year salary  =  22000(1 + 3%)3

=  22000(1.03)3

=  24040

Question 3 :

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb−c × yc−a × za−b = 1 .

Solution :

Since a, b and c are in A.P,

b - a  = c - b  = d (common difference)

We need to prove,

xb−c × yc−a × za−b = 1

Let us try to convert the powers in terms of one variable.

2b  =  c + a - a + a

2b  =  c - a + 2a

2(b - a)  =  c - a

2d  =  c - a

If c - b  =  d, then b - c = -d

If b - a  =  d, then a - b = -d

L.H.S

xb−c × yc−a × za−b  =  x−d × y2d × zd   ---(1)

y = √xz

By applying the value of y in (1)

=  x−d × (√xz)2d × zd

=  x−d × (xz)d × zd

=  x−d + d  z-d + d

=  1

Hence proved. After having gone through the stuff given above, we hope that the students would have understood, "Word Problems in Geometric Sequence".

Apart from the stuff given in this section "Word Problems in Geometric Sequence"if you need any other stuff in math, please use our google custom search here.