Question

An employee gets Rs 300 per month in his 11th year of service and got Rs 495 per month in his 24th year of service. If his monthly salary is in A.P his initial salary and his incriment are
1.Rs 200,Rs 10 . 2.Rs300,Rs 10 . 3.Rs150,Rs 15 ​

Answer 1

Answer:

Rs.15

Step-by-step explanation:

See the pic,I have explained the problem in the pic.

I hope it helps.

Answer 2

Concept: AP stands for Arithmetic Progression.

$a_{n}$=a+(n-1)d

Given:

$a_{11} =300\\a_{24} =495$

Find:

Solution: Let a is initial salary and d is incriment.

$a_{11} =300$

a+10d=300→(1)

$a_{24} =495$

a+23d=495→(2)

Subtraction of eq.(1) from (2)

a+23d-(a+10d)=495-300

a+23d-a-10d=195

13d=195

d=15

Substitute the value of d in eq.(1)

a+10X15=300

a=300-150

a=150

Final answer: His initial salary was Rs.150.

His incriment per year was Rs.15.

#SPJ3