**Frequently Asked Questions on Chapter 5 – Arithmetic Progressions**

**In which of the following situations, does the list of numbers involved make an arithmetic progression and why? What is the taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km?**

We can write the given condition as; Taxi fare for 1 km = 15 Taxi fare for first 2 kms = 15 + 8 = 23 Taxi fare for first 3 kms = 23 + 8 = 31 Taxi fare for first 4 kms = 31 + 8 = 39 And so on……Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.

**Choose the correct choice in the following and justify 30 ^{th} term of the A.P: 10, 7, 4, … is 97, 77,−77,−87?**

A.P. = 10, 7, 4, … Therefore, we can find, First term, *a* = 10 Common difference, *d* = *a*_{2} − *a*_{1 }= 7 − 10 = −3 As we know, for an A.P.*a _{n}* =

*a*+ (

*n*− 1)

*d*

Putting the values;

*a*

_{30}= 10 + (30 − 1) (−3)

*a*

_{30}= 10 + (29) (−3)

*a*

_{30}= 10 − 87 = −77 Hence, the correct answer is option C.

**Which term of the A.P. 3, 8, 13, 18, … is 78?**

Given the A.P. series as3, 8, 13, 18, … Thus,*First term, a* = 3*Common difference, d* = *a*_{2} − *a*_{1} = 8 − 3 = 5 Let the *n*^{th} term of given A.P. be 78. Now as we know,*a _{n}* =

*a*+ (

*n*− 1)

*d*

*Therefore,*

78 = 3 + (

*n*− 1) 5 75 = (

*n*− 1) 5 (

*n*− 1) = 15

*n*= 16 Hence, 16

^{th}term of this A.P. is 78.

**Find the sum of the following APs 2, 7, 12 ,…., to 10 terms.**

Given, 2, 7, 12 ,…, to 10 terms For this A.P., first term, a = 2 And common difference, d = a_{2} − a_{1} = 7 − 2 = 5 n = 10 We know that,the formula for sum of nth term in AP series is,*S _{n}* =

*n*/2 [2a + (

*n*– 1)

*d*]

*S*= 10/2 [2(2) + (10 – 1) × 5] 5[4 + (9) × (5)] 5 × 49 = 245

_{10}**Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0]**

Given the AP series is 121, 117, 113, . . ., Thus, first term, a = 121 Common difference, d = 117-121= -4 By the nth term formula,*a _{n}* =

*a*+ (

*n*− 1)

*d*

Therefore, a

_{n}= 121 + (n − 1)(-4) = 121-4n + 4 =125-4n To find the first negative term of the series,

*a*< 0 Therefore, 125-4n < 0 125 < 4n n>125/4 n>31.25 Therefore, the first negative term of the series is 32

_{n }^{nd}term.

**Write the first four terms of the A.P. when the first term a = 10 and the common difference d = 10 are given**

Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …

a1 = a = 10

a2 = a1+d = 10+10 = 20

a3 = a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

a5 = a4+d = 40+10 = 50

And so on…

Therefore, the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of this A.P. will be 10, 20, 30, and 40.

For the following A.P. 3, 1, – 1, – 3 …, write the first term and the common difference

3, 1, – 1, – 3 …

First term, a = 3

Common difference, d = Second term – First term

1 – 3 = -2

d = -2

Therefore, a = 3, d = -2

Which term of the A.P. 3, 8, 13, 18, … is 78?

Given the A.P. series as 3, 8, 13, 18, …

First term, a = 3

Common difference, d = a2 − a1 = 8 − 3 = 5

Let the nth term of given A.P. be 78. Now as we know,

an = a + (n−1)d

Therefore,

78 = 3+(n −1)5

75 = (n−1)5

(n−1) = 15

n = 16

Hence, the 16th term of this A.P. is 78.