The hand-borewell driller charges Rs. $200$ for the first one metre only and raises drilling charges at the rate of Rs. $30$ for every subsequent metre. Write a progression for the above data.

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Solution

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 $R s . 200$ and $R s . 30$ for subsequent metre.
therefore $a = 200$ and $d = 30$

So progression $t_{n} = 200 + (n - 1) 30$ where $n = 1, 2, 3 . . .$

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The hand-borewell driller charges Rs.200 for the first one metre only and raises drilling charges at the rate of Rs.30 for every subsequent metre. Write a progression for the above data.

For progression first term a and common difference d is required.Progression will be tn=a+(n−1)dAs for the first one metre charges are Rs.200 and Rs.30 for subsequent metre.therefore a=200 and d=30So progression tn=200+(n−1)30 where n=1,2,3...

The hand-borewell driller charges Rs. $200$ for the first one metre only and raises drilling charges at the rate of Rs. $30$ for every subsequent metre. Write a progression for the above data.