Question

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

Answer 1

Heya friend,

Here is your answer,

Given, salary = Rs 8000

annual increment = Rs 100

Money earned in first year = 8000 x 12 = Rs 96000

Smilarly , money earned in second year with increment of rs100 = 8100 x 12 = Rs 97200

Now for third year = 8200 x 12 = rs 98400

here a = Rs 96000

d = Rs1200

n = 10yrs

Now we can see that it becomes an AP (Arethmatic Progression)
Therefore using the formula,

Sn = n/2 [( 2a + (n-1) × (d)]

=> 5 ( 192000 + 10800 )

=> 5 ( 202800)

=> Rs 1014000 (ans)


Hope it helps you.
Thank you.

Answer 2

here's solution

Given, salary = 8000

annual increment = 100

Money earned in first year = 8000 x 12 = Rs 96000

Smilarly , money earned in second year with increment of rs100 = 8100 x 12 = Rs 97200

Now for third year = 8200 x 12 =  rs 98400

here a = Rs 96000

d = Rs1200

n = 10yrs

using  formula 

SN = n/2 [ ( 2a +(n-1) (d) ]

=> 5 (  192000 + 10800 )

=> 5 ( 202800)

=> Rs1014000

I hope it will help u :)