Question

A woman takes up a job of Rs.8,000 per month with a monthly increment of Rs.100. What will she earn over a period of 10 years?

Answer 1

Hint: In this question, we first need to look into some basic definitions of series. Then from the definition of arithmetic progression by using the formula for sum of n terms of an arithmetic progression we can find the earning over a period of 10 years.
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Complete step by step solution:
ARITHMETIC PROGRESSION (AP):
A sequence in which the difference of two consecutive terms is constant, is called Arithmetic Progression (AP).
$a,a+d,a+2d$ is an AP, where a is the first term and d is the common difference.
$n\text{th}$ Term of an AP: If a is the first term, d is the common difference and l is the last term of an AP,
$n\text{th}$ term is given by:
$l={{a}_{n}}=a+\left( n-1 \right)d$
Common difference of an AP:
$d={{T}_{n}}-{{T}_{n-1}}$
Sum of n terms of an AP is given by:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Now, from the conditions given in the question we get,

& a=8000 \\\ & d=100 \\\ \end{aligned}$$ Conversion of years to months 1 year = 12 months $$\begin{aligned} & \Rightarrow 10years=10\times 12 \\\ & \therefore 10years=120months \\\ \end{aligned}$$ $$n=120$$ $$\begin{aligned} & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\\ & \Rightarrow {{S}_{n}}=\dfrac{120}{2}\left[ 2\times 8000+\left( 120-1 \right)100 \right] \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow {{S}_{n}}=60\left[ 16000+\left( 119\times 100 \right) \right] \\\ & \Rightarrow {{S}_{n}}=60\left[ 16000+11900 \right] \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow {{S}_{n}}=60\times 27900 \\\ & \therefore {{S}_{n}}=16,74,000 \\\ \end{aligned}$$ Hence, she will earn Rs.1674000 over a period of 10 years. Note: If a constant is added or subtracted from each term of an AP, then the resulting sequence is an AP with the same common difference. If the terms of an AP are chosen at regular intervals, then they form an AP. Instead of using the arithmetic progression and using the sum of n terms in an AP series we can also solve this problem algebraically. But, for that we need to add 100 for every month till the end of 10 years which is a lengthy process and there are chances that we may neglect any of the terms in between which changes the result completely. While substituting the respective terms in the formula of sum of n terms of an arithmetic progression we need to first convert years into months because the increment in the salary is at the end of every month.