A person is employed in a company at rupees 3000 per month and he would get an increase of rupees hundred per year find the total amount which he received in 25 year and the monthly salary in the last year?

Question

Answer

Solution:

Given,
Monthly salary = Rs. 3000
Increase per year = Rs. 100

To find:
1. Total amount received in 25 years
2. Monthly salary in the last year

Calculation:
1. Total amount received in 25 years:
- The person receives an increase of Rs. 100 per year.
- So, the salary after n years will be: 3000 + 100n
- After 25 years, the salary will be: 3000 + 100 x 25 = Rs. 5500
- Now, we can use the formula for the sum of an arithmetic progression to find the total salary received in 25 years.
- S = n/2 [2a + (n-1)d], where a = first term, d = common difference, n = number of terms
- Here, a = 3000, d = 100, n = 25
- S = 25/2 [2 x 3000 + (25-1) x 100]
- S = 25/2 [6000 + 2400]
- S = 25 x 4200
- S = Rs. 1,05,000

2. Monthly salary in the last year:
- We already found that the salary after 25 years is Rs. 5500.
- So, the monthly salary in the last year is Rs. 5500/12 = Rs. 458.33 (approx.)

Therefore, the total amount received in 25 years is Rs. 1,05,000 and the monthly salary in the last year is Rs. 458.33 (approx.).

Answer

So per monthly salary is = Rs 3,000 and he would get an increase of Rs 100 per year
so per year increase = 100 ×12 = 1,200
1st year he would get = 3,000×12=36,000
2nd year = 36,000 +1,200=37,200
3rd year = 37,200+ 1,200= 38,400
4th year = 38,400+1,200= 39,600
so....
36,000 , 37,200 , 38,400 , 39,600...

a=36,000
d= 37,200-36,000=1200
n = 25

so he required find total amount which he received in 25 year
and monthly salary in the last year

Sn= n/2 [2a+(n-1)×d]
Sn= 25/2 [2×36,000+(25-1)×1,200]
= 25/2 [ 72,000+24×1,200]
=25/2 [ 72,000+28,800]
=25/2 ×1,00,800
=25 ×50,400
Sn= 12,60,000

tn=a+(n-1)×d
= 36,000+(25-1)×1,200
= 36,000+24×1,200
= 36,000 + 28,800
tn= 64,800

so monthly salary in the last year
64,800÷12 =5,400

SO I HOPE YOU UNDERSTAND THIS STEP BY STEP ANSWER
( this question is on page no 6.15 illustration no 1 in icai module)
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Answer

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