Advertisements
Advertisements
प्रश्न
The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?
उत्तर
Given that, the students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school.
Given that, the number of flags = 27 and distance between each flag = 2 m.
Also, the flags are stored at the position of the middle most flag i.e., 14th flag and Ruchi was given the responsibility of placing the flags.
Ruchi kept her books, where the flags were stored i.e., 14th flag and she coluld carry only one flag at a time.
Let she placed 13 flags into her left position from middle most flag i.e., 14th flag.
For placing second flag and return her initial position distance travelled = 2 + 2 = 4 m.
Similarly, for placing third flag and return her initial position,
Distance travelled = 4 + 4 = 8 m
For placing fourth flag and return her initial position,
Distance travelled = 6 + 6 = 12 m
For placing fourteenth flag and return her initial position,
Distance travelled = 26 + 26 = 52 m
Proceed same manner into her right position from middle most flag i.e., 14th flag.
Total distance travelled in that case = 52 m
Also, when Ruchi placed the last flag she return his middle most position and collect her books.
This distance also included in placed the last flag.
So, these distance form a series.
4 + 8 + 12 + 16 + ... + 52 ...[For left]
And 4 + 8 + 12 + 16 + ... + 52 ...[For right]
∴ Total distance covered by Ruchi for placing these flags
= 2 × (4 + 8 + 12 + ... + 52)
= `2 xx [13/2 {2 xx 4 + (13 - 1) xx (8 - 4)}]` ...`{{:(∵ "Sum of n terms of an AP")/(S_n = n/2[2a + (n - 1)d]):}}`
= `2 xx [13/2 (8 + 12 xx 4)]` ...[∵ Both sides of number of flags i.e., n = 13]
= 2 × [13(4 + 12 × 2)]
= 2 × 13(4 + 24)
= 2 × 13 × 28
= 728 m
Hence, the required distance is 728 m in which she did cover in completing this job and returning back to collect her books.
Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the 14th flag in her left position or 27th flag in her right position
= (2 + 2 + 2 + ... + 13 times)
= 2 × 13
= 26 m
Hence, the required maximum distance she travelled carrying a flag is 26 m.
APPEARS IN
संबंधित प्रश्न
If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero
If the 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
Which term of AP 72,68,64,60,… is 0?
Which term of the AP `20, 19 1/4 , 18 1/2 , 17 3/4 ` ,..... is the first negative term?
The angles of quadrilateral are in whose AP common difference is 10° . Find the angles.
What is the 5th term form the end of the AP 2, 7, 12, …., 47?
The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Q.10
The sum of the first 2n terms of the AP: 2, 5, 8, …. is equal to sum of the first n terms of the AP: 57, 59, 61, … then n is equal to ______.
Find the sum of all even numbers from 1 to 250.