Question

Suppose you work at a bowling alley. After work one day, you decide to line up bowling pins in a triangular
pattern with one pin in the first row, two pins in the second, three pins in the third, and so on.
(a)
How many total pins would you need to use in order to complete four rows?
(b) How many total pins would you need to use in order to complete ten rows?
(c) How many total pins would you need to use in order to complete one hundred rows?
How about one thousand rows?

Answer 1

1) We are given that every row has one extra pin than the last one and since there are four rows and the First row start with 1 pin then the total number of pins in four rows would be- 1+2+3+4= 10

2) To complete 10 rows the total number of pins needed would be-  

Applying arithmetic progression to find out Sn = n/2 (2a + (n-1) d) = 5 x (2+9) =55

3) Now for 100 rows applying same formula again Sn = n/2 (2a + (n-1) d) = 50 x (2+99) =5050

4) Now for 1000 rows applying the formula Sn = n/2 (2a + (n-1) d) = 500x (2+999) =500500  

Answer 2

Answer:

The total pins to complete four rows is 10.

The total number of pins to complete 10 rows is 55.

The total number of pins to complete 100 rows is 5050.

The total number of pins to complete 1000 rows is 500500.

Explanation:

(a)

For fours rows, the arrangement will be like;

1 pin in one row, 2 pins in second row, 3 pins in third row and 4 pins in fourth row. Then total pins is,

$n= 1+2+3+4 = 10$

Thus, the total pins to complete four rows is 10.

(b)

For 10 rows, apply the formula for sum of arithmetic progressions as,

$S_{n}=\dfrac{n}{2}(2a+(n-1)d)$

Here, a is first term and d is common difference.

$S_{10}=\dfrac{10}{2}(2 \times 1+(10-1) \times 1)\\S_{10}=5(2 +9)\\S_{10}=55$

Thus, the total number of pins to complete 10 rows is 55.

(c)

For 100 rows, apply the formula as,

$S_{n}=\dfrac{n}{2}(2a+(n-1)d)$

$S_{100}=\dfrac{100}{2}(2 \times 1+(100-1) \times 1)\\S_{100}=50(2 +99)\\S_{100}=5050$

Thus, the total number of pins to complete 100 rows is 5050.

(d)

For 1000 rows, apply the formula as,

$S_{n}=\dfrac{n}{2}(2a+(n-1)d)$

$S_{1000}=\dfrac{1000}{2}(2 \times 1+(1000-1) \times 1)\\S_{1000}=500(2 +999)\\S_{1000}=500500$

Thus, the total number of pins to complete 1000 rows is 500500.

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