Math, asked by harikrishna53, 11 months ago

.
The hand-bore well dealer charges Rs 200.- for the first one metre only and raises drilling
charges at the rate of rupees 30/- for every subsequent metre. Write a progression for the
above data ​

Answers

Answered by rahul123437
3

A progression for the  above data = 200,230,260,290...............

Given:

The hand-bore well dealer charges Rs 200.

For the first one meter only and raises drilling .

Charges at the rate of rupees 30/- for every subsequent meter.

To find:

A progression for the  above data.

​Explanation:

For first the hand-bore well dealer charges Rs 200.

Charges at the rate of rupees 30/- for every subsequent meter.

That is for second meter charge is equal to rupees 30/- more than first meter.

Second meter charge = 200+30 = 230

Third meter charge is equal to rupees 30/- more than second meter.

Third meter charge = 230+30 = 260 Rs

So on......

A progression for the  above data = 200,230,260,290...............

This is arithmetic progression with common difference = 30

To learn more...

1)Sum of three consecutive terms of an Arithmetic Progression is 42 and their product is 2520.

https://brainly.in/question/11255431

2)Measures of sides of a triangle are in Arithmetic progression. Its perimeter is 30 cm, and the

between the longest and shortest side is 4 cm, then find the meqasure of the sides

https://brainly.in/question/13742618

Answered by bindhu4989
0

Answer:

For progression first term a and common difference d is required.

Progression will be t

n

=a+(n−1)d

As for the first one metre charges are Rs.200 and Rs.30 for subsequent metre.

therefore a=200 and d=30

So progression t

n

=200+(n−1)30 where n=1,2,3..

hope it helps

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