Question

In a potato race, 20 potatoes are placed in a line of intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?; In a potato race, 20 potatoes are placed in a line of intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Answer 1

AP: (2×24)+2×(24+4)+2×(24+4+4).........

AP: 48,56,64,........

We have,

No. of terms ,n =20

Common difference,d=(56-48)=8

$a_{n} = a + (n - 1)d$

$a_{20} = 48 + (20 - 1)8 \\ \: \: \: \: \: = 48 + 19 \times 8 \\ \: \: \: \: \: = 48 + 152 \\ \: \: \: \: = 200$

So the last term of the AP is 200

$s_{n} = \frac{n}{2} (a + a_{n})$

$s_{20} = \frac{20}{2} (48 + 200) \\ \: \: \: \: \: = 10(248) \\ \: \: \: \: = 2480$

So the total distance travelled is 2480 m

Answer 2

Solution -:

Given -:;

➣in patato race potato are placed = 20

➣first potato starting point from 24 m

Find -:

➣how far would he run in bringing back all the potatoes

Required Answer-:

➣2480 m

Let's Begin -:

According to the given condition the sequence become 24 , 28 ... 20th term

here a =24 D =28 -24 n = 20

$sn \: = \frac{n}{2} (2a + (n - 1)d \:$

$⇢ s _{20} = \frac{20}{2} (2 \times 24 + (20 - 4)4) \\ \\ ⇢ 10(48 + 76) = 10 \times 124 = 1240$

So the contestant run to bring the potatoes back to the starting place

$⇢ 2s _{n} = 2 \times 1240 \\ \\⇢ = 2480m \:$

➣2480 for would he run in bringing back all the patatoes.