Question 13: Find the sum of the first 15 multiples of 8.

**Solution:** We have a = 8, d = 8 and n = 15

Sum of first 15 multiples of 8 can be calculated as follows:

`S=n/2[2a+(n-1)d]`

`=(15)/(2)(2xx8+14xx8)`

`=15(8+7xx8)`

`=15xx64=960`

Question 14: Find the sum of the odd numbers between 0 and 50.

**Solution:** There are 25 odd numbers between 0 and 50

Sum of first n odd numbers = n^{2}.

`= 25^2 = 625`

Question 15: A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

**Solution:** Here; a = 200, d = 50 and n = 30

We can find the penalty by using the sum of 30 terms;

`S=n/2[2a+(n-1)d]`

`=(30)/(2)(2xx200+29xx50)`

`=15(400+1450)`

`=15xx1850=Rs. 27750`

Question 16: A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

**Solution:** Here; d = 20 n = 7 and S_{n} = 700

We know;

`S=n/2[2a+(n-1)d]`

Or, `700=7/2(2a+6xx20)`

Or, `700=7(a+60)`

Or, `a+60=700÷7=100`

Or, `a=100-60=40`

So the prizes in ascending order are as follows: Rs. 40, Rs. 60, Rs. 80, Rs. 100, Rs. 120, Rs. 140 and Rs. 160

Question 17: In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

**Solution:** Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1.

Similarly,

No. of trees planted by 3 sections of class 2 = 6

No. of trees planted by 3 sections of class 3 = 9

No. of trees planted by 3 sections of class 4 = 12

We have; a = 3, d = 3 and n = 12

We can find the total number of trees as follows:

`S=n/2[2a+(n-1)d]`

`=(12)/(2)(2xx3+11xx3)`

`=6(6+33)`

`=6xx39=234`

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