Question

A woman takes up a job of Rs 8000 per month with a increment of Rs100 . What will she earn over a period of 10 years ?

Answer 1

Tn=a+ ( n-1)×d
tn= 8000+ (10-1)×100
tn=8000+900
.•.tn= 8900

Answer 2

current salary : ₹8000
Annual salary in 1st year = 12 * 8000 = 96000
Increment every year = ₹100
Annual salary in 2nd year = 8100 * 12 = 97200
in 3rd year - 98400

For the given A. P.
a = 96000 d = 1200
n = 10 years

now
Sn = n/2{2a + (n - 1)d}
S10 = 10/2(2*96000 + 9*1200)
= 5 (19200 + 10800)
= 5 (202800)
= 1014000 ₹

:)