Not the exact question you were looking for?
Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.
Arithmetic Progressions
EXERCISE 5.3
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
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 (see Fig. 5.6). Fig. 5.6 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 × 5 + 2 × (5 + 3)] 71
Step-by-Step Solution
Key Concept: The distances of the potatoes from the bucket form an arithmetic progression. The total distance run is twice the sum of these distances because the competitor goes to each potato and returns to the bucket.
1. Distances of the potatoes from the bucket\
The first potato is $5\,\text{m}$ away. Each subsequent potato is $3\,\text{m}$ farther. Hence the distances form an A.P. with\
\[ a = 5,\quad d = 3,\quad n = 10. \]
The $k^{\text{th}}$ potato is at distance \[ D_k = a + (k-1)d = 5 + (k-1)\times 3 \] meters.
2. Sum of all distances\
Using the sum formula for an A.P.:\
\[ S = \frac{n}{2}\bigl[2a + (n-1)d\bigr] \]
Substituting the values:\
\[ S = \frac{10}{2}\bigl[2\times5 + 9\times3\bigr] = 5\bigl[10 + 27\bigr] = 5\times37 = 185\ \text{m}. \]
So the total of the one‑way distances to all potatoes is $185\,\text{m}$.
3. Total distance run\
For each potato the competitor runs to it and back, i.e., twice the one‑way distance. Hence\
\[ \text{Total distance} = 2 \times S = 2 \times 185 = 370\ \text{m}. \]
4. Verification with the hint\
For the first two potatoes: $2\times5 + 2\times(5+3) = 10 + 16 = 26\,\text{m}$, which matches the method above. Extending the same reasoning to all ten potatoes gives $370\,\text{m}$.
Thus the competitor has to run a total of $370$ metres.
The first potato is $5\,\text{m}$ away. Each subsequent potato is $3\,\text{m}$ farther. Hence the distances form an A.P. with\
\[ a = 5,\quad d = 3,\quad n = 10. \]
The $k^{\text{th}}$ potato is at distance \[ D_k = a + (k-1)d = 5 + (k-1)\times 3 \] meters.
2. Sum of all distances\
Using the sum formula for an A.P.:\
\[ S = \frac{n}{2}\bigl[2a + (n-1)d\bigr] \]
Substituting the values:\
\[ S = \frac{10}{2}\bigl[2\times5 + 9\times3\bigr] = 5\bigl[10 + 27\bigr] = 5\times37 = 185\ \text{m}. \]
So the total of the one‑way distances to all potatoes is $185\,\text{m}$.
3. Total distance run\
For each potato the competitor runs to it and back, i.e., twice the one‑way distance. Hence\
\[ \text{Total distance} = 2 \times S = 2 \times 185 = 370\ \text{m}. \]
4. Verification with the hint\
For the first two potatoes: $2\times5 + 2\times(5+3) = 10 + 16 = 26\,\text{m}$, which matches the method above. Extending the same reasoning to all ten potatoes gives $370\,\text{m}$.
Thus the competitor has to run a total of $370$ metres.
Correct Answer: 370 m
Mathbee AI Mentor (Free Demo)
Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.