Ch 5: Arithmetic Progressions (Part 1)
Master the basics of AP and the nth Term Formula [ an = a + (n-1)d ] with step-by-step NCERT solutions.
Exercise 5.1 Solutions
Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
Fare for 1st km (a₁) = ₹15
Fare for 2nd km (a₂) = 15 + 8 = ₹23
Fare for 3rd km (a₃) = 23 + 8 = ₹31
Difference (d) = a₂ - a₁ = 23 - 15 = 8.
Difference (d) = a₃ - a₂ = 31 - 23 = 8.
Since the difference between consecutive terms is constant (8), Yes, it forms an AP.
Q2. Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) a = 10, d = 10
Solution:
First term (a₁) = a = 10
Second term (a₂) = a₁ + d = 10 + 10 = 20
Third term (a₃) = a₂ + d = 20 + 10 = 30
Fourth term (a₄) = a₃ + d = 30 + 10 = 40
The first four terms are: 10, 20, 30, 40.
Q4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, ...
a₂ - a₁ = 4 - 2 = 2
a₃ - a₂ = 8 - 4 = 4
Since a₂ - a₁ ≠ a₃ - a₂, It is not an AP.
(v) 3, 3+√2, 3+2√2, 3+3√2, ...
a₂ - a₁ = (3 + √2) - 3 = √2
a₃ - a₂ = (3 + 2√2) - (3 + √2) = √2
Difference is constant (d = √2). Yes, it forms an AP.
Next 3 terms: (3+4√2), (3+5√2), (3+6√2).
Exercise 5.2 (nth Term Formula)
💡 Master Formula: an = a + (n - 1)d
Q4. Which term of the AP: 3, 8, 13, 18, ... is 78?
Solution:
Given AP: 3, 8, 13, 18...
First term (a) = 3
Common difference (d) = 8 - 3 = 5
Let the nth term be 78. So, an = 78.
Using formula: an = a + (n - 1)d
78 = 3 + (n - 1)5
78 - 3 = (n - 1)5
75 = 5(n - 1)
15 = n - 1 ⇒ n = 16
Therefore, the 16th term of this AP is 78.
Q11. Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
Solution:
Here, a = 3, d = 15 - 3 = 12.
First, find the 54th term (a₅₄):
a₅₄ = a + 53d = 3 + 53(12) = 3 + 636 = 639
We need to find 'n' such that an = a₅₄ + 132
an = 639 + 132 = 771
Now, 771 = a + (n - 1)d
771 = 3 + (n - 1)12
768 = 12(n - 1)
64 = n - 1 ⇒ n = 65
The 65th term will be 132 more than the 54th term.
Q17. Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.
Solution (Reverse AP Trick):
To find a term from the last, simply reverse the AP.
Original AP: 3, 8, 13, ... 253 (Here d = 5).
Reversed AP: 253, 248, 243, ... 3.
For the reversed AP:
First term (A) = 253
Common difference (D) = -5 (Sign changes when reversed)
Now, find the 20th term of this new AP:
A₂₀ = A + 19D
A₂₀ = 253 + 19(-5)
A₂₀ = 253 - 95 = 158
The 20th term from the last is 158.
Important PYQs
Q. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
Given: a₃ = 4 ⇒ a + 2d = 4 --- (Eq 1)
Given: a₉ = -8 ⇒ a + 8d = -8 --- (Eq 2)
Subtracting Eq 1 from Eq 2:
(a + 8d) - (a + 2d) = -8 - 4
6d = -12 ⇒ d = -2
Putting d = -2 in Eq 1:
a + 2(-2) = 4 ⇒ a - 4 = 4 ⇒ a = 8
We need to find n where an = 0:
a + (n - 1)d = 0
8 + (n - 1)(-2) = 0
(n - 1)(-2) = -8
n - 1 = 4 ⇒ n = 5
Therefore, the 5th term is zero.