Maths

100 MATHEMATICS
(ix) 1, 3, 9, 27, . . . (x) a, 2a, 3a, 4a, . . .
(xi) a, a 2 , a 3 , a 4 , . . . (xii) 2, 8, 18 , 32, . . .
(xiii) 3, 6, 9 , 12 , . . . (xiv) 1 2 , 3 2 , 5 2 , 7 2 , . . .
(xv) 1 2 , 5 2 , 7 2 , 73, . . .
5.3 nth Term of an AP
Let us consider the situation again, given in Section 5.1 in which Reena applied for a
job and got selected. She has been offered the job with a starting monthly salary of
Rs 8000, with an annual increment of Rs 500. What would be her monthly salary for
the fifth year?
To answer this, let us first see what her monthly salary for the second year
would be.
It would be Rs (8000 + 500) = Rs 8500. In the same way, we can find the monthly
salary for the 3rd, 4th and 5th year by adding Rs 500 to the salary of the previous year.
So, the salary for the 3rd year = Rs (8500 + 500)
= Rs (8000 + 500 + 500)
= Rs (8000 + 2 × 500)
= Rs [8000 + (3 – 1) × 500] (for the 3rd year)
= Rs 9000
Salary for the 4th year = Rs (9000 + 500)
= Rs (8000 + 500 + 500 + 500)
= Rs (8000 + 3 × 500)
= Rs [8000 + (4 – 1) × 500] (for the 4th year)
= Rs 9500
Salary for the 5th year = Rs (9500 + 500)
= Rs (8000+500+500+500 + 500)
= Rs (8000 + 4 × 500)
= Rs [8000 + (5 – 1) × 500] (for the 5th year)
= Rs 10000
Observe that we are getting a list of numbers
8000, 8500, 9000, 9500, 10000, . . .
These numbers are in AP. (Why?)