Name: Problems on General Term of an A.P
Uploaded: 2019-07-19
Duration: 5 min 14 s
Description: In this video, we are going to learn how to find the nth term of an AP and an example problem to learn how to find the nth term of an arithmetic progression.

Question

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 $4^{th}$ year with respect to the offers $A$ and $B$ ?

Accepted Answer

Here, the salaries offered by the company in both the offers $A$ and $B$ in different years are in GP.

We know that the $n^{th}$ term of a GP is given by $a_{n} = a r^{n - 1}$ , where $a$ is the first term and $r$ is the common ratio of the GP.

Offer $A$

Here, the starting salary is $₹ 20, 000$ and the annual increase in salary is $A$ .

So, in this case, the first term $a_{A} = ₹ 20000$ and $r_{A} = (1 + \frac{6}{100}) = 1.06$

So, the salary offered by the company in the $4^{th}$ year is given by $a_{4 A} = a_{A} {r_{A}}^{3} = ₹ 20000 \times {(1.06)}^{3} = ₹ 23820.32$ .

Offer $B$

Here, the starting salary is $₹ 22, 000$ and the annual increase in salary is $3 %$ .

So, in this case, the first term $a_{B} = ₹ 22000$ and $r_{B} = (1 + \frac{3}{100}) = 1.03$

So, the salary offered by the company in the $4^{th}$ year is given by $a_{4 B} = a_{B} {r_{B}}^{3} = ₹ 22000 \times {(1.03)}^{3} = ₹ 24039.994 ≃ ₹ 24039.99$

Therefore, Sivamani's salary in the $4^{th}$ year for offer $A$ is $₹ 23820.32$ and that for offer $B$ is $₹ 24039.99$ .

Important Questions on Numbers And Sequences

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