![]() |
VOOZH | about |
An arithmetic progression (AP), also known as an arithmetic sequence, is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is called the "common difference."
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,
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.
Question 2: If the 4th term of an AP is 8 with a common difference of 2. Find out the arithmetic progression up to 8 terms.
Solution:
Given that, the fourth term, a4 is 8 and the common difference is 2,
So the fourth term can be written as,
a + (4 - 1) Ć 2 = 8 [a = first term]
= a + 3 Ć 2 = 8
= a = 8 - 3 Ć 2
= a = 8 - 6
= a = 2So the first term a1 is 2,
Now, a2 = a1 + 2 = 2 + 2 = 4 similarly the series upto 8 term is = 2, 4, 6, 8, 10, 12, 14, 16.
Question 3: Find the 40th term in the 2, 5, 8, 11, 14,... sequence
Solution:
a = 2
d = 5 - 2 = 3
n = 40an = a + ( n - 1 )d
a40 = 2 + ( 40 - 1 )3
= 2 + 117
a40 = 119
Question 4: Find the 9th term in the 1/3, 2/3, 1, 4/3... sequence
Solution:
a = 1/3
d = 2/3 - 1/3 = 1/3
n = 9an = a + ( n -1 )d
a9 = 1/3 + ( 9 - 1 )(1/3)
= 1/3 + 8/3
= 9/3 = 3
Question 5: Find the 15th term of the sequence 9, 7, 5, 3, 1,...
Solution:
a = 9
d = 7 - 9 = 5 - 7 = -2
n = 15a15 = a + ( n - 1 ) Ć d
= 9 + ( 15 - 1 ) Ć (-2)
= 9 + ( 14 ) Ć ( -2 )
= 9 - 28
a15 = -19Hence, 15th term in the sequence 9,7,5,3,1,... is -19
Question 6: Find the general term of the series 4,7, 10,13......
Solution:
In the given AP, we have
a1 = 4, d = 3So, general term of the given AP is
an = a1 + (n - 1)d
an = 4 + (n - 1)3
an = 4 + 3n - 3
an = 3n + 1General term = 3n + 1.
Question 7: Find the 50th term of the sequence 293, 290, 287, 284, 281,...
Solution:
a = 293
d = 290 - 293 = 287 - 290 = -3
n = 50a50 = a + (n - 1) Ć d
= 293 + ( 50 - 1 ) Ć (-3)
= 293 + ( 49 ) Ć ( -3 )
= 293 - 147
a50 = 146Hence, 50th term in the sequence 293, 290, 287, 284, 281,... is 146
Question 8: Find the 100th term of the sequence 100, 50, 0, -50, -100,...
Solution:
a = 100
d = 50 - 100 = 0 - 50 = -50
n = 100a100 = a + ( n - 1 ) Ć d
= 100 + ( 100 - 1 ) Ć (-50)
= 100 + ( 99 ) Ć ( -50 )
= 100 - 4950
a100 = -4850Hence, 100th term in the sequence 100, 50, 0, -50, -100,... is -4850
Solution.
First odd natural number is 1
Common Difference is 2So, 20th 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 Ć 40
Sum = 400So, sum of first 20 odd numbers = 400
Solution:
In the given AP,
a1 = 30
d = -3
an = 0Now, to find n
an = a1 + (n-1)d
0 = 30 + (n - 1)(-3)
0 = 30 - 3n + 3
3n = 33
n = 11So, the 11th term is 0
Solution:
In the given AP, we have
a1 = 36
d = -6To find the first negative term,
36 + (nā1)(ā6) < 0
36 + 6 - 6n < 0
42 - 6n < 0
-6n < - 42
n > 7So, the 8th term is the first negative term of the given AP.
Solution:
So, the given common difference = 4.
Now, we want to calculate a20 - a15,
So,
a20 = a + 19d
a15 = a + 14dThen,
a20 - a15 = a+ 19d - a - 14d
a20 - a15 = 5d
a20 - a15 = 5 Ć 4 = 20So the value of a20 - a15 is 20.
Solution:
In the given AP, we have following values
a = 3
d = 5Let the unknown term is n term,
Now,
an = a20 + 55
a1 + (n - 1)d = a1 + 19d + 55
3 + (n-1)d = 3 + 19Ć5 + 55
5n - 5 = 95 + 55
5n = 155
n = 31So, the 31st term is 55 more than the 20th term.
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 Ć 12b + (18 - 1)(-4b)]
Sum = 9[24b - 68b]
Sum = 9 Ć (-44b)
Sum = -396bSo sum of first 18 terms is -396b.
Question 15: If the 3rd term of an arithmetic sequence is 12 and the 7th term is 24, what is the first term?
Solution:
The nth term : an = a+ (n-1) d
a3 = 12 = a + 2d - (1)
a7 = 24 = a + 6d - (2)Subtract (1) from (2), (2) - (1) :
a7 - a3 = 24 - 12 = 12 = a + 6d - a - 2d = 6d - 2d = 4d
12 = 4d
d = 12 / 4 = 3Put value of d in (1) :
12 = a + 2(3)
a = 12 - 6 = 6Hence the 1st term of arithmetic sequence is 6.
Question 1: In an AP, if d = 8, n = 10, and an = 80, then Find the value of a.
Question 2: Find the 19th term of the AP: 5, 11, 17, 23, ...?
Question 3: What is the sum of first 30 even natural numbers?
Question 4: Which term of the arithmetic progression 13, 18, 23, ..... is 80 more than its 20th term?
Question 5: If the sum of first p terms of an A.P., is ap2 + bp, find its common difference.
Question 6: The first term of an A.P. is p and its common difference is q. Find its 10th term
Question 7: 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?
Question 8: How many terms of the A.P. 18, 16, 14, ⦠⦠be taken so that their sum is zero?
Question 9: 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
Question 10: Find the First Term Given a Condition:
Answer key
- a = 8
- 19th term of the AP: 113
- Sum of first 30 even natural numbers: 930
- 36th term is 80 more than the 20th term.
- Common difference d = 2a
- 10th term of the AP: p + 9q
- 18th term is 34d
- 19 terms should be taken.
- Proof: 25th term = 3 Ć 11th term
- Middle term: 9
Middle term: 27