A man accepts a position with an initial salary of Rs 5200 per month.

Given the man’s salary in first month is 5200rs and then it increases every month by 320rs

Hence the sequence of his salary per month will be 5200, 5200 + 320, 5200 + 640…

The sequence is in AP with first term as a = 5200,

Common difference as d = 320

a) Now we have to find salary in 10th month that is 10th term of the AP

The nth term of AP is given by tn = a + (n – 1) d

Where a is the first term and d is the common difference

We have to find t10

⇒ t10 = 5200 + (10 – 1) (320)

⇒ t10 = 5200 + 2880

⇒ t10 = 8080

Hence salary in 10th month is 8080rs

b) To get the total earnings in first year we have to add first 12 terms of the sequence that is we have to find S12

The sum of first n terms of AP is given by Sn = n/2 (2a + (n – 1) d)

Where a is the first term and d is common difference

S12 = (12/2) (2 (5200) + (12 – 1) 320)

⇒ S12 = 6(10400 + 11(320))

⇒ S12 = 6(10400 + 3520)

⇒ S12 = 6(13920)

⇒ S12 = 83520

Hence his total earnings in first year is 83520rs