A man accepts a position with an initial salary of Rs 5200 per month. It is understood that he will receive an automatic increase of Rs 320 in the very next month and each month thereafter.
(a) Find his salary for the tenth month
(b) What is his total earnings during the first year?
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
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