Table of contents  
What are Arithmetic Progressions?  
Infinite and Finite Arithmetic Progression (AP)  
Solved Examples Related to AP  
nth term of an AP  
Sum of first n terms of an AP 
Patterns can be observed in our daily lives, even in something as simple as a savings account. For example, if we start with Rs. 2000 and add Rs. 500 each month, the balance follows a predictable pattern: Rs. 2000, Rs. 2500, Rs. 3000, and so on. This is called an Arithmetic progression (AP).
In this Class 10 document, we will learn the basics of APs, their properties, practical uses, and how to solve problems related to them.
Arithmetic Progressions (APs) are sequences of numbers in which the difference between consecutive terms remains constant. This constant difference is called the common difference. In an arithmetic progression, each term can be obtained by adding the common difference to the previous term.
Here is a list of numbers, observe the pattern of the numbers.
Arithmetic Progressions (APs) can be categorized as either finite or infinite based on the number of terms they contain.
Understanding the distinction between finite and infinite APs helps us analyze different types of sequences and apply appropriate methods and formulas accordingly.
Example 1: In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
a) The cost of digging a well after every meter of digging when it costs Rs. 150 for the first meter and rises by Rs. 50 for each subsequent meter.
Sol:
Cost of digging the first meter = Rs.150
Cost of digging second metre = Rs.(150 + 50) = Rs.200
Cost of digging third metre = Rs.(150 + 2× 50) = Rs. 250
In this case, each term is obtained by adding Rs. 50 to the preceding term. Hence, they make an AP.
b) The amount of water present in a cylinder when a vacuum pump removes 1⁄4 of the air remaining in the cylinder at a time.
Sol:
Let the amount of air present in the cylinder be x units.
According to the question,
Amount of air left in the cylinder after using vacuum pump first time
Amount of air left in the cylinder after using vacuum pump the second time=
List of numbers
As the common difference of the terms is not the same, they do not form AP.
Example 2: Find the common difference of the AP
Sol:
Common difference (d) = Second Term – First Term
Therefore, common difference = 2
Example 3: For what value of k will k + 10 , 2k, and 2k + 8 are the consecutive terms of an AP.
Sol:
If k + 10 , 2k, and 2k + 8 are in AP then,
a_{1} = k + 10 , a_{2} = 2k and a_{3} = 2k + 8
Common Difference (d)
Example 4: If the numbers 2n − 2, 3n + 1, and 6n − 2 are in AP, then find n and hence find the numbers.
Sol:
The numbers 2n − 2, 3n + 1, and 6n − 2 are in AP then,
a_{1} = 2n − 2 , a_{2} = 3n + 1 and a_{3} = 6n − 2
Common difference (d) = a_{2} − a_{1} = a_{3} − a_{2}
d = 3n + 1 − (2n − 2) = 6n − 2 − (3n + 1)
3n + 1 − 2n + 2 = 6n − 2 − 3n − 1
n + 3 = 3n − 3 ⇒ 3n − n = 3 + 3
2n = 6 ⇒ n = 3
Therefore the value of n is 3.
The numbers are,
2n − 2 = 2 × 3 − 2 = 6 − 2 = 4
3n + 1 = 3 × 3 + 1 = 9 + 1 = 10
6n − 2 = 6 × 3 − 2 = 16
Then,
The n^{th} term an of the AP with the first term a and common difference d is given by
a_{n} = a + (n − 1)d
Example 5: Find the 25th term of the AP: 5, 5/2, 0, 5/2
Sol:
Here, a = −5
Common Difference (d) = a_{2} − a_{1}
We know, a_{n }= a + (n − 1)d
25^{th} term, a_{25}
25th term of the given AP is 55.
Example 6: For an AP, if a_{18} − a_{14} = 36, then find the common difference d.
Sol:
We know, a_{n} = a + (n − 1)d
a_{18} = a + (18 − 1)d = a + 17d
a_{14} = a + (14 − 1)d = a + 13d
a_{18} − a_{14} = 36
a + 17d − (a + 13d) ⇒ a + 17d − a − 13d
4d = 36
d = 9
Therefore, the common difference (d) = 9
Example 7: If a_{n} = 6 − 11n, then find the common difference.
Sol:
Given: a_{n }= 6 − 11n ............... (1)
Replacing n by n+1 in eq (1) we get
a_{(n+1) }= 6 – 11(n+1)
a_{(n+1)} = 6 – 11n11
a_{(n+1)} =  11n – 5
Common difference, d = a_{(n+1}) – a_{n}
d= 11n5 – (611n)
d = 11
Example 8: Find the 7th term of the sequence whose nth term is given by an = (−1)^{n−1}. n^{2}
Sol:
Given: an = (−1)^{n−1}. n^{2}
The 7th term of the sequence, a_{7} = (−1)^{7−1}. 7^{2}
= (−1)^{6}. 7^{2} = 49
Example 9: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Sol:
Let a be the first term and d be the common difference of the given AP.
Given: a_{3 }= 12 and a_{50} = 106
We know, a_{n} = a + (n − 1)d
a_{3} = a + (3 − 1)d ⇒ a_{3} = a + 2d
a_{3 }= a + 2d = 12→Eq 1
a_{50} = a + (50 − 1)d ⇒ a_{50} = a + 49d
a_{50} = a + 49d = 106→Eq 2
Subtracting Eq 1 from Eq 2
a + 49d − (a + 2d) = 106 − 12
a + 49d − a − 2d = 94
47d = 94 ⇒ d = 2
Putting the value of d in Eq 1 we get,
a + 2 × 2 = 12
a = 12 − 4 = 8
29^{th} term, a_{29} = a + 28d = 8 + 28 × 2 = 8 + 56 = 64
Example 10: Find how many twodigit numbers are divisible by 7.
Sol:
Twodigit numbers are 10, 11, 12, 13, ..................97, 98, 99, 100.
Here 14, 21, 28...................... 91, 98 are divisible by 7.
This list of numbers forms an AP, where a = 14 and
d = 21 − 14 = 7
Let the number of terms be n, then an = 98
98 = 14 + (n − 1)7 ⇒ 98 = 14 + 7n − 7
7n + 7 = 98 ⇒ 7n = 98 − 7
7n = 91
n = 13
Hence, 13 twodigit numbers are divisible by 7.
Example 11: If the nth term of an AP is (2n + 2), find the sum of first n terms of the AP.
We have,
a_{n} = 2n + 2 ⇒ a1 = 2 × 1 + 2 = 4
Therefore, a_{1 }= a is the first term and a_{n} = l is the last term of the AP.
As we know the first and the last term of the AP, the sum of n terms is given by,
Sum of first n terms of the given AP is n[n + 3]
Example 12: Find the sum of the first 25 terms of an AP, whose nth term is given by a_{n} = 6 − 3n.
Given: a_{n} = 6 − 3n
a_{1} = 6 − 3 × 1 = 3a_{25} = 6 − 3 × 25 = −69
Therefore, a_{1} = a = 3 is the first term and a_{25} = l = −69
Therefore, the sum of the first 25 terms of the given AP is 825.
Example 13: Find the sum of all twodigit odd positive numbers.
Twodigit odd positive numbers are 11, 13, 15..................99 which form an AP.
Here, First term a = 11, last term (l) = 99 and common difference (d) = 13 – 11 = 2
Now we have to find the number of terms.We know, l = a_{n} = a + (n − 1)d
99 = 11 + (n − 1) × 2
99 = 11 + 2n − 2 ⇒ 99 = 9 + 2n
99 − 9 = 2n ⇒ 90 = 2n
Therefore, the sum of all twodigit odd positive numbers is 2475.
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