Arithmetic Progression is an important topic for students of class 9 and class 10. It is used in our everyday lives with some real-life applications. So, to learn them and understand them is essential to ease out our several tasks.
Important Formulas on Arithmetic Progression
Various formulas on arithmetic progression are:
|
Nth term of an AP |
an = a1 + (n-1)d |
|---|---|
|
Sum of first n terms of an AP |
Sn = n/2(a1 + an) Sn = n/2[2a1 + (n-1)d] |
|
Nth Term from Sum of an AP |
an = Sn – Sn-1 |
|
Number of Terms in an AP |
n = (an – a1)/d + 1 |
where,
- an is nth term
- a1 is first term
- d is common difference
- Sn is sum of the first n terms
- Sn-1 is sum of the first n-1 terms
- n is number of terms
Practice Questions on Arithmetic Progression
Q 1. In an AP, if d = 8, n = 10 and an = 80, then Find the value of a?
Q 2. Find the 19th term of the AP: 5, 11, 17, 23, …?
Q 3. What is the sum of first 30 even natural numbers?
Q 4. Which term of the arithmetic progression 13, 18, 23, ….. is 80 more than its 20th term?
Q 5. If the sum of first p terms of an A.P., is ap2 + bp, find its common difference.
Q 6. The first term of an A.P. is p and its common difference is q. Find its 10th term
Q 7. Find the 17th term from the end of the AP: 1, 6, 11, 16 … … 211, 216.
Q 8. If seven times the 7th term of an A.P. is equal to eleven times the 11th term, then what will be its 18th term?
Q 9. How many terms of the A.P. 18, 16, 14, … … be taken so that their sum is zero?
Q 10. 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
Practice Questions on Arithmetic Progression with Solution
Q 1. What is the sum of first 20 odd natural numbers ?
Solution.
First odd natural number is 1
Common Difference is 2
So, twentieth term is
a20 = a1 + (n-1)d
a20 = 1 + (20-1)2
a20 = 1 + 38
a20 = 39
So,
Sum = n/2(a1 + a20)
Sum = 20/2(1 + 39)
Sum = 10 × 40
Sum = 400
So, sum of first 20 odd numbers = 400
Q 2. Find the common difference from the given AP: 1, 6, 11, 16, 21.
Solution:
To find common difference we need to subtract consecutive terms
d = an – an-1
d = 6 – 1 = 5
So, common difference of the given AP is 5.
Q 3. Find the general term of the series 4,7, 10,13……
Solution:
In the given AP, we have
a1 = 4, d = 3
So, general term of the given AP is
an = a1 + (n-1)d
an = 4 + (n-1)3
an = 4 + 3n – 3
an = 3n + 1
General term = 3n + 1.
Q 4. Which term of the arithmetic progression 30, 27, 24,…. is 0?
Solution:
In the given AP,
- a1 = 30
- d = -3
- an = 0
Now, to find n
an = a1 + (n-1)d
0 = 30 + (n-1)(-3)
0 = 30 – 3n + 3
3n = 33
n = 11
So, the 11th term is 0
Q 5. Find the first negative term of the arithmetic progression 36, 30, 24,…..?
Solution:
In the given AP, we have
- a1 = 36
- d = -6
To find the first negative term,
36 + (n−1)(−6) < 0
36 + 6 – 6n < 0
42 – 6n < 0
-6n < – 42
n > 7
So, the 8th term is the first negative term of the given AP.
Q 6. If the common difference of an A.P. is 4, then 𝑎20 − 𝑎15 is
Solution:
So, the given common difference = 4.
Now, we want to calculate a20 – a15,
So,
a20 = a + 19d
a15 = a + 14d
Then,
a20 – a15 = a+ 19d – a – 14d
a20 – a15 = 5d
a20 – a15 = 5 × 4 = 20
So the value of a20 – a15 is 20.
Q 7. Find the middle term of the A.P. 6, 13, 20, … , 216.
Solution:
In the given AP,
- First term a1 = 6
- Last term an = 216
- Common difference = 7
Now, to find the number of terms, n = (an – a1)/d + 1
n = (216 – 6)/7 + 1
n = 210/7 + 1
n = 30 + 1
n = 31
So, middle term is (n + 1) / 2
= (31 + 1)/2
= 16
Now to calculate the middle term
a16 = a1 + 15×d
a16 = 6 + 15 × 7
a16 = 6 + 105 = 111
So the middle term of the given AP is 111
Q 8. In an AP, if 𝑑 = −2, 𝑛 = 5 and 𝑎𝑛 = 0, then find the value of 𝑎 ?
Solution:
Given values are:
- d = -2
- n = 5
- an = 0
So, to calculate follow these steps:
an = a + (n-1)d
0 = a + (5-1)(-2)
0 = a – 8
a = 8
Q 9. Which term of the arithmetic progression 3, 8, 13, ….. is 55 more than its 20th term?
Solution:
In the given AP, we have following values
- a = 3
- d = 5
Let the unknown term is n term,
Now,
an = a20 + 55
a1 + (n-1)d = a1 + 19d + 55
3 + (n-1)d = 3 + 19×5 + 55
5n – 5 = 95 + 55
5n = 155
n = 31
So, the 31st term is 55 more than the 20th term.
Q 10. Determine the sum of the first 18 terms, of the arithmetic progression 12b, 8b, 4b,……
Solution:
In the given AP we have,
- a1 = 12b
- d = -4b
So, the sum of first 18 terms is
Sum = n/2[2a1 + (n-1)d]
Sum = 9[2×12b + (18-1)(-4b)]
Sum = 9[24b – 68b]
Sum = 9 × (-44b)
Sum = -396b
So sum of first 18 terms is -396b.
FAQs on Arithmetic Progression
Who is known as father of Arithmetic?
Brahmagupta is known as father of Arithmetic
What is sum of first 5 natural numbers?
- a1 = 1
- d = 1
sum = n/2[2a1 + (n-1)d]
sum = 5/2 × [2×1 + (5-1)d]
sum = 5/2 × [2 + 4] = 5×3 = 15
So, the sum of first 5 natural numbers is 15.
Find the common difference in the following AP: 1, 3, 5, …
Common difference in the following AP: 1, 3, 5, … is 2
Write Formula to find the sum of first n terms
Sum of n terms = n/2[2a1 + (n-1)d]
Can sum of an AP be zero.
Yes, sum of an AP may be zero sometimes.