**Q) In a school examination hall, the teacher makes students sit in such a way that no students can cheat from other student. So, the teacher decides to mark the numbers on each chair from 1. 2. 3……. **

**If there are 26 students and each student is seated at alternate position in examination hall such that the sequence formed is 1, 3, 5. Based on the above information, answer the following questions:**

**i. Find the common difference between two seats.**

**ii. hat type of sequence is formed, to follow the seating arrangement of students in the examination hall?**

**iii a. Find the seat number of the last student in the examination hall.**

**OR**

**iii b. Find the seat number of 10th vacant seat in the examination hall.**

**Ans: **It is given that the seats are numbered in sequence and 26 students are being seated on alternate seats.

Let’s start solving the questions.

**i.** It is given that 1^{st} student is given seat no. 1, next student gets seat no. 3, next student gets seat no. 5 and so on…..

Hence, the difference between first 2 consecutive seats is: 3 -1 = 2

Similarly, the difference between next 2 seats is: 5-3 = 2.

Since this difference is common across the seats, hence it is common difference.

**Hence, Common difference between two seats is 2.**

**ii. ** Since Students are being seated at seat nos. 1, 3, 5, ….. with a common difference of 2.

**This sequence is an** **AP of all odd integers. **It has following characteristics:

# Its first term, a = 1

# Its common difference, d = 2

# This AP will have total 26 terms which is equal to no. of students.

**iii a.** Since last student is 26^{th} student, we need to find the seat no. of this student which will be the value of 26^{th} term of this AP.

In an AP, n^{th} term T_{n} = a + (n – 1) d

∴ T_{26} = 1 + (26 – 1) (2) = 1 + 25 x 2 = 51

**Therefore, last student will sit in 51 no. seat.**

*[ Note for Students: This AP is of odd integers and hence value of each term will be odd. You need to recheck your solution, if your answer is coming even no.]*

**OR**

**iii b.** We can see that 1^{st} student sits on 1^{st} seat, then 2^{nd} seat goes vacant, 2^{nd} student comes on 3^{rd} seat and then 4^{th} seat goes vacant and so on…

It means, 1^{st} vacant seat seat comes after 1^{st} student, 2^{nd} vacant seat comes after 2^{nd} student and so on…

Hence, 10^{th} vacant seat will come after 10^{th} student.

Therefore, if we find out seat number of 10^{th} student, we need to just add 1 to get seat no. of our 10^{th} vacant seat.

Now, seat number of 10^{th} student will be value of 10^{th} term of our AP.

We know that, in an AP, n^{th }term T_{n} = a + (n – 1) d

∴ T_{9} = 1 + (10 – 1) (2) = 1 + 9 x 2 = 19

10^{th} Vacant seat’s no. will be just after this i.e. 19 + 1 = 20

**Hence, number of 10 ^{th} vacant seat is 20.**

*[ Note for Students: This number is next to students seat and will have to be an even number. Recheck your solution, if your answer is coming odd no.]*

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