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". 

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