Question

a woman takes up a job of Rs 8000 per month with a monthly increment of ₹100. What will she earn over a period of 10years?

Answer 1

$\bf\huge\boxed{\boxed{\boxed{Cybary\:Radhe\:Radhe}}}$

Monthly salary = Rs. 8000

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

Hence for 3rd year :-

= 8200 × 12

= Rs. 98400

Hence :-

a = Rs. 96000

d = Rs. 1200

n = 10 years

Sn =$\bf\huge\frac{n}{2}$ {2a + (n - 1)d}

S₁₀ = $\bf\huge\frac{10}{2}${2 × 9600 + (10 - 1)1200}

= 5 (192000 + 10800)

= 5 × 102800

$\bf\huge\boxed{\boxed{\boxed{= \:Rs.\: 1014000}}}$

$\bf\huge\boxed{\boxed{\boxed{Radhe\:Radhe}}}$