Question

a woman takes up a job of Rs 8000 per month with an annual increment of Rs 100 .what will she earn over a period of 10 years? is the answer 1014000

Answer 1

Solution:-
Monthly salary = Rs. 8000
Then money earned in the 1st year = 8000*12 = Rs. 96000
Annual increment = Rs. 100
Money earned in the 2nd year with increment = 8100*12 = Rs. 97200
Now for 3rd year = 8200*12 = Rs. 98400
So, here 
a = Rs. 96000
d = Rs. 1200
n = 10 years
Sn = n/2{2a + (n - 1)d}
S₁₀ = 10/2{2*9600 + (10 - 1)1200}
= 5{192000 + 10800}
= 5*102800
= Rs. 1014000
Answer

Answer 2

Answer: 1014000

Step-by-step explanation is given below in the photo