Question

22. If a worker in a factory receives one rupee on the first day from the second days onwards his wage is increased by one rupee every day. What is the total amount of wage he receives after 40 days?

Answer 1

Answer:

RS.820

Step-by-step explanation:

sum of n natural number = n(n+1)/2 = 40(41)/2 = Rs.820/-48. A bell in a clock rings once at 1 O'clock, twice at 2 O'clock, thrice at 3 O'clock andso on.. Then how many times it rings in a day. Correct Option : D)sum of n natural number = n(n+1)/2 = 12(13)/2 = 78 = 2 times (78) = 15649. What is the sum of natural numbers between 20 and 100.

Answer 2

Answer:

The worker will get 820 rupees after 40 days.

Step-by-step explanation:

Given,a worker in a factory receives one rupee on the first day and from second day he receives his wage is increased by one rupee every day.

According to question,

Firstday the worker get 1 rupees.

Second day the worker get (1+1) = 2 rupees

Third day the worker get (2+1) = 3 rupees.

Fourth day the worker get (3+1) = 4 rupees.

By this way,we can say on 40 number day,the worker will get 40 rupees.

So, series of his wages is 1,2,3,.....,40.

We know,

If 1,2,3,....,n is a series then sum of the series is $\frac{n(n + 1)}{2}$

In this question n = 40.

So,sum of his total wages after 40 days

$\frac{40(40 + 1)}{2 } \\ = \frac{40 \times 41}{2} \\ = 20 \times 41 \\ = 820$

Therefore after 40 days the worker will get 820 rupees.

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