Find the total distance covered after seven jumps.

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Q.
Computer Animations: The animation on a new computer game initially allows the hero of the game to jump a (screen) distance of 10 inch over booby traps and obstacles. Each successive jump is limited to 3/4 inch less than the previous one.

Based on the above information, answer the following questions:

(ii) Find the total distance covered after seven jumps.

1 $54 \frac{1}{2} i n c h$

2 $54 \frac{1}{4} i n c h$

3 $54 \frac{2}{1} i n c h$

4 $- 54 \frac{1}{2} i n c h$

Ans.
(2)

Total distance covered after seven jumps is given by

$S_{7} = 7 / 2 2 \times 10 + (7 - 1) \times (- 3 / 4) [∵ S_{n} = n / 2 (2 a + (n - 1) d)] \Rightarrow S_{7} = 7 / 2 (20 - 6 \times 3 / 4) = 7 / 2 (20 - 9 / 2) = 7 / 2 \times 31 / 2 = 217 / 4 = 54 1 / 4 i n c h . "$

Solution

1 $54 \frac{1}{2} i n c h$

2 $54 \frac{1}{4} i n c h$

3 $54 \frac{2}{1} i n c h$

4 $- 54 \frac{1}{2} i n c h$

Ans.
(2)

Total distance covered after seven jumps is given by

$S_{7} = 7 / 2 2 \times 10 + (7 - 1) \times (- 3 / 4) [∵ S_{n} = n / 2 (2 a + (n - 1) d)] \Rightarrow S_{7} = 7 / 2 (20 - 6 \times 3 / 4) = 7 / 2 (20 - 9 / 2) = 7 / 2 \times 31 / 2 = 217 / 4 = 54 1 / 4 i n c h . "$

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Solution

1 5412 inch

2 5414 inch

3 5421 inch

4 -5412 inch

Ans. (2) Total distance covered after seven jumps is given by S7=7/22×10+(7-1)×(-3/4)[∵Sn=n/2(2a+(n-1)d)]⇒S7=7/2(20-6×3/4)=7/2(20-9/2)=7/2×31/2=217/4=54 1/4 inch."

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1 $54 \frac{1}{2} i n c h$

2 $54 \frac{1}{4} i n c h$

3 $54 \frac{2}{1} i n c h$

4 $- 54 \frac{1}{2} i n c h$

Ans.
(2)

Total distance covered after seven jumps is given by

$S_{7} = 7 / 2 2 \times 10 + (7 - 1) \times (- 3 / 4) [∵ S_{n} = n / 2 (2 a + (n - 1) d)] \Rightarrow S_{7} = 7 / 2 (20 - 6 \times 3 / 4) = 7 / 2 (20 - 9 / 2) = 7 / 2 \times 31 / 2 = 217 / 4 = 54 1 / 4 i n c h . "$