Question

Aman participates in an orange race. In the race, 20 oranges are placed in a line of intervals of 4 meters with the first orange 24 meters from the starting point. Aman is required to bring the oranges back to the starting place one at a time. How far would he run in bringing back all the oranges?

Answer 1

Answer:

Step-by-step explanation:

Since every orange is placed at a difference of 4 meters and the first potato is placed at 24 meters from the starting position. Every orange is placed at 24m, 28m, 32m,  ….20 terms.

Now to bring ever orange one at a time, Aman needs to cover the double of the distance = 48, 56,  …20 terms.

So putting the values in the sum of AP formula, a = 48, d= 8, n = 20.

Total distance travel = 20/2 [2 * 48 + (20-1)*8]

= 2480 meters

Answer 2

The distance run in bringing back all the oranges one at a time to the starting point would be equal to 2480 meters.

Given:

Number of oranges = 20

The distance of the first orange from the starting point = 24 meters

The separation between each orange = 4 meters

To Find:

The distance run by him while bringing back all the oranges.

Solution:

→ The first orange is kept at a distance of 24 meters from the starting point.

→ The distance between each orange is 4 meters.

• Distance run in bringing the first orange (D₁) = 2 × 24 = 48 meters
• Distance run in bringing the second orange (D₂) = 48 + 8 = 56 meters
• Distance run in bringing the third orange (D₃) = 56 + 8 = 64 meters
• Distance run in bringing the fourth orange (D₄) = 64 + 8 = 72 meters

→ As we can see that the above sequence D₁, D₂, D₃, D₄....so on, is an Arithmetic Progression (A.P.).

→ The first term of this sequence being D₁ equal to 48 meters and common difference (d) being equal to 8 meters.

→ The general nth term of this A.P. (Dₙ) would be equal to D₁ + (n - 1)d.

• Distance run in bringing the nth orange (Dₙ) = [D₁ + (n - 1)d] meters

∴ Distance run in bringing the 20th orange (D₂₀) = 48 + (20 - 1)(8) = 200 m

→ We would add these individual distances run i.e. D₁ + D₂ +D₃ +D₄....so on, in order to get the total distance run.

→ Sum of 'n' terms of this A.P. (Sₙ) would be given as (n/2)[D₁ + (n - 1)d]

                                  $\boldsymbol{\therefore S_{n} =\frac{n}{2} [2S_{1} +(n-1)d]}$

∴ To calculate the distance run in bringing back the 20 oranges, we would add the first 20 terms of this A.P. that is D₁ + D₂ + D₃ + D₄ +....+ D₂₀.

                                  $\boldsymbol{\therefore S_{n} =\frac{n}{2} [2S_{1} +(n-1)d]}\\\\\boldsymbol{\therefore S_{20} =\frac{20}{2} [2(48) +(20-1)8]}\\\\\boldsymbol{\therefore S_{20} =(10) [96 +152]}\\\\\boldsymbol{\therefore S_{20} =(10)\times(248)}\\\\\boldsymbol{\therefore S_{20} =2480\:meters}$

Therefore the distance run in bringing back all the oranges one at a time to the starting point would be equal to 2480 meters.

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