Arithmetic Progression successive semicircles NCERT Exercise 5.3 part eight Class Ten Mathematics

# Arithmetic Progression

## Exercise 5.3 Part 8

Question 18: A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?

[Hint : Length of successive semicircles is l1, l2, l3, l4, . . . with centres at A, B, A, B, . . ., respectively.]

Solution: Circumference of first semicircle = πr = 0.5π

Circumference of second semicircle = πr = 1π = π

Circumference of third semicircle =πr = 1.5π

It is clear that a = 0.5 π, d = 0.5π and n = 13

Hence; length of spiral can be calculated as follows:

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

=(13)/(2)(2xx0.5π+12xx0.5π)

=(13)/(2)xx7π

=(13)/(2)xx7xx(22)/(7)=143 cm

Question 19: 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution: We have; a = 20, d = - 1 and Sn = 200

We know;

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

Or, 200=n/2[2xx20+(n-1)(-1)]

Or, 400=n(40-n+1)

Or, 400=41n-n^2

Or, n^2-41n+400=0

Or, n^2-25n-16n+400=0

Or, n(n-25)-16(n-25)=0

Or, (n-16)(n-25)=0

Thus, n = 16 and n = 25

If number of rows is 25 then;

a_25 = 20 + 24 xx (- 1)

= 20 – 24 = - 4

Since; negative value for number of logs is not possible hence; number of rows = 16

a_16 = 20 + 15 xx (- 1)

= 20 – 15 = 5

Thus, number of rows = 16 and number of logs in top rows = 5

Question 20: In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 xx 5 + [2 xx (5 + 3)]

Solution: Distance covered in picking and dropping the 1st potato = 2 xx 5 = 10  m

Distance covered in picking and dropping the 2nd potato = 2 xx (5+3) = 16  m

Distance covered in picking and dropping the 3rd potato = 2 xx (5 + 3 + 3) = 22  m

So, we have a = 10, d = 6 and n = 10

Total distance can be calculated as follows:

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

=(10)/(2)(2xx10+9xx6)

=5(20+54)

=5xx74=370 m