**Word Problems in Geometric Sequence :**

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

**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)^{5 }

= 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 x^{b−c} × y^{c−a} × z^{a−b} = 1 .

**Solution :**

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

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

We need to prove,

x^{b−c} × y^{c−a} × z^{a−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

x^{b−c} × y^{c−a} × z^{a−b } = x^{−d} × y^{2d} × z^{−}^{d }---(1)

y = √xz

By applying the value of y in (1)

= x^{−d} × (√xz)^{2d} × z^{−}^{d }

= x^{−d} × (xz)^{d} × z^{−}^{d }

= 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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