Advertisements
Advertisements
प्रश्न
An insect is on the 0 point of a number line, hopping towards 1. She covers half the distance from her current location to 1 with each hop. So, she will be at `1/2` after one hop, `3/4` after two hops, and so on.

- Make a table showing the insect’s location for the first 10 hops.
- Where will the insect be after n hops?
- Will the insect ever get to 1? Explain.
Advertisements
उत्तर
a. On the basis of given information in the question, we can arrange the following table which shows the insect’s location for the first 10 hops.
| Number of hops | Distance covered | Distance left | Distance covered |
| 1. | `1/2` | `1/2` | `1 - 1/2` |
| 2. | `1/2(1/2) + 1/2` | `1/4` | `1 - 1/4` |
| 3. | `1/2(1/4) + 3/4` | `1/8` | `1 - 1/8` |
| 4. | `1/2(1/8) + 7/8` | `1/16` | `1 - 1/16` |
| 5. | `1/2(1/16) + 15/16` | `1/32` | `1 - 1/32` |
| 6. | `1/2(1/32) + 31/32` | `1/64` | `1 - 1/64` |
| 7. | `1/2(1/64) + 63/64` | `1/128` | `1 - 1/128` |
| 8. | `1/2(1/128) + 127/128` | `1/256` | `1 - 1/256` |
| 9. | `1/2(1/256) + 255/256` | `1/512` | `1 - 1/512` |
| 10. | `1/2(1/512) + 511/512` | `1/1024` | `1 - 1/1024` |
b. If we see the distance covered in each hops
Distance covered in 1st hop = `1 - 1/2`
Distance covered in 2nd hops = `1 - 1/4`
Distance covered in 3rd hops = `1 - 1/8`

Distance covered in n hops = `1 - (1/2)^n`
c. No, because for reaching `1, (1/2)^n` has to be zero for some finite n which is not possible.
APPEARS IN
संबंधित प्रश्न
Simplify and write the answer in the exponential form.
`1/8 xx (3)^(-3)`
The value for (–7)6 ÷ 76 is ______.
ap × bq = (ab)pq
Solve the following:
100–10
A new born bear weighs 4 kg. How many kilograms might a five year old bear weigh if its weight increases by the power of 2 in 5 years?
Find a single repeater machine that will do the same work as hook-up.

Find a single repeater machine that will do the same work as hook-up.

Find x.
2x + 2x + 2x = 192
If a = – 1, b = 2, then find the value of the following:
ab × ba
Simplify:
`((3^-2)^2 xx (5^2)^-3 xx (t^-3)^2)/((3^-2)^5 xx (5^3)^-2 xx (t^-4)^3`
