One of the most important and exam oriented question from Chapter name- Arithmetic Progression

Class 10th

Chapter number- 9

Exercise :- 9.6

This type of question has been asked in previous years exams.

In this question we have been given that the students have to fix the flags on the straight passage of the school.

They have 27 flags to be fixed at intervals of every 2 meter. 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.

It is the condition that She could carry only one flag at a time.

Now we have to find out how much distance did she cover in completing this job and returning back to collect her books.

Also we have to calculateÂ the maximum distance she travelled carrying a flag.

This is very conceptual and important question and we have to solve it by using the arithmetic progression based concept.

CBSE DHANPAT RAI publication

CBSE Mathematics Class 10th

Question 74

Ruchi has to fix 13 flags to the left of the middle position, 1 flag at the middle position and

remaining 26 flags at the right of the middle position.

Distance covered in fixing 1st flag to the left of the middle position = 2 + 2 = 4m

Distance covered in fixing 2nd flag to the left of the middle position = 4 + 4 = 8m

Distance covered in fixing 3rd flag to the left of the middle position = 8 + 8 = 16m

So, first term(a) = 4, common difference(d) = 8 â€“ 4 = 4, number of terms = 13

and last term(a

_{13}) = 26+26 = 52mSo we use the sum of n terms of an A.P.: S

_{n}Â = n[2a + (n â€“ 1)d] / 2.So, Distance covered in fixing flags to the left of the middle position = Sum of 13 terms of the A.P.

= 13[2(4) + (13 â€“ 1)4] / 2

= 13[8 + 48]/2

= 13[28]

= 364

Total distance covered = 2 Ã— 364 = 728m

Hence, the maximum distance Ruchi travelled carrying a flag

would be the distance she travelled while carrying the last flag

to the left or to the right, which in our case is 26m.