Q) A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs. 15,000 after 4 years of service and Rs. 18,000 after 10 years of service, what was his starting salary and what was the annual increment?

Ans:

Step 1: Let’s consider the starting salary was X and annual increment was Y.

Since the increment is annual and it is a fixed amount, it means the salary will increase by a fixed amount every year.

Hence, this forms an AP where the first term is X and the common difference is Y

Step 2: Now, let’s calculate the salary after 4 years of service

This will be given by 5th term i.e. n = 5 and value of this 5th term by T_n = a + (n – 1) d

∴ T₅ = X + (5 – 1) Y

Given that T5 = 15000

∴ X + 4 Y = 15000 ….(i)

Step 3: Now, let’s calculate the salary after 10 years of service

This will be given by 11th term i.e. n = 11 and value of this 11th term T_n = a + (n – 1) d

∴ T₁₁ = X + (11 – 1) Y

Given that T11 = 18000

∴ X + 10 Y = 18000 ….(ii)

Step 4: By subtracting equation (i) from equation (ii), we get:

(X + 10 Y) – (X + 4 Y) = 18000 – 15000

∴ 10 Y – 4 Y = 3000

∴ 6 Y = 3000

∴ Y =

∴Y = 500

Step 5: By substituting the value of Y in equation (i), we get:

X + 4 Y = 15000

∴ X + 4 (500) = 15000

∴ X + 2000 = 15000

∴ X = 15000 – 2000 = 13000

Therefore, the starting salary was Rs. 13,000 and the annual increment was Rs. 500.

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