In series and sequence, arithmetic, geometric and harmonic sequences are important concepts. A set of numbers which follows a certain pattern is called a sequence, for example, 3, 6, 9, 12, ……….(a sequence of multiple of 3), also even numbers, odd numbers, follow a particular pattern. The numbers are named as terms of a sequence. In this article, we come across Arithmetic progression solved examples and definition.
Arithmetic Progression Definition
A sequence of numbers in which the difference between the two consecutive numbers is always constant can be termed as arithmetic progression. The difference is called the common difference.
- The first term of an AP is denoted as “a”.
- The common difference is represented as “d”.
- The common difference can either be positive or negative.
A sequence containing finite terms is called a finite sequence whereas a sequence with infinite terms, that is without the last term can be defined as an infinite sequence.
Properties of an arithmetic progression
1.If we add a constant to each term of an A.P, the resulting sequence will be also an A.P.
2.If we subtract a constant from each term of an A.P, the resulting sequence will be also an A.P.
3.If we multiply a constant to each term of an A.P, the resulting sequence will be also an A.P.
4.If each term of an A.P is divided by a non zero constant, the resulting sequence will be also an A.P.
We use the following notations for an A.P
a= first term
d = common difference
l = last term
n = no. of terms
Sn = sum of n terms of an A.P
Arithmetic Progression Formula
Formula | Terms | |
The general formula for finding the n^{th} term of an AP | \(\begin{array}{l}T_n = a + (n-1)d\end{array} \) |
Where d = a_{n} – a_{n-1} and T_{n} is the n^{th} term of the sequence. |
The formula for finding the sum to n terms of an AP | \(\begin{array}{l}S_n = \frac{n}{2} [2a + (n-1)d]\end{array} \) |
where S_{n }is the sum to n terms, a is the first term and d is the common difference |
When the last term is known, then the sum of n terms | \(\begin{array}{l}S_n = \frac{n}{2} [a + l]\end{array} \) |
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Solved Examples on Arithmetic Progression
Example 1: Let the b^{th} term of an A.P. be q, and c^{th} term be p, then find its d^{th} term.
Solution:
T_{b }= a + (b − 1)d = q …..(i) and
T_{c }= a + (c − 1)d = p …… (ii)
From (i) and (ii),
we get d = −(b−c)/(b−c) = −1
Putting the value of d in equation (i), then a = b + c − 1
Now, d^{th} term is given by A.P.
T_{d }= a + (d − 1)d
= (b + c − 1) + (d − 1)(−1)
= b + c − d
Example 2: Write the sum of the numbers which are divisible by 8 between 100 and 1000.
Solution:
The series will be 104, 112, ………………….., 992
Common difference = 8
n = 992/8 – 96/8 = 124 – 12 = 112
Hence the required sum is,
S_{n} = 112/2 (104 + 992) = 61376
Example 3: If the 8^{th} term of an AP is zero, then what is the ratio of its 28^{th} and 18^{th} term?
Solution:
Given that 8th term = a+(8−1)d = 0
⇒ a+7d = 0
Now ratio of 28^{th} and 18^{th} terms
(a + 27d) /(a + 17d) = {(a + 7d) + 20d} / {(a + 7d) + 10d} = 20d/10d = 2/1
Example 4: If tan (mθ) = tan (nθ), in which sequence will the different values of θ be?
Solution:
We have tan (mθ) = tan(nθ)
⇒ mθ = Mπ + (nθ)
⇒ θ = Mπm − n, putting M = 1, 2, 3………, we get
π/m−n, 2π/m−n, 3π/m−n,……… which are obviously in A.P.
Since common difference d = π/m − n
Example 5: What will be the n^{th} term of the series 3.9 + 6.12 + 9.15 + 12.18 + …..?
Solution:
Given series 3.9+6.12+9.15+12 .18+…..
First factors are 3, 6, 9, 12, whose n^{th }term is 3n and the second factors are 9, 12, 15, 18
t_{n }= [9 + (n − 1)3] = (3n + 6)
Hence n^{th }term of given series = 3n(3n + 6).
Example 6: If 3x, x + 9, 6x + 1 are in A.P., then what will be the value of x?
Solution:
3x, x + 9, 6x + 1 are in A.P.
Therefore, (x + 9) = (3x) + (6x + 1)/2 = 9x + 1 / 2
⇒ 2x + 18 = 9x + 1
⇒ 7x = 17
⇒ x = 2.42
Example 7: Find the sum of all natural numbers lying between 101 and 199, which are multiples of 5.
Solution:
We know the nearest number to 101, which is a multiple of 5 and lies between 101 and 199, is 105. Similarly, 195 is a multiple of 5, which lies between 101 and 199. So we take the last term of A.P as 195.
We have first term, a = 105
Common difference, d = 5
Last term, l = 195
l = a + (n – 1) d .
195 = 105 + (n – 1)5
195 – 105 = (n – 1)5
5n – 5 = 90
5n = 90 + 5 = 95
n = 95/5 = 19
Sum of n terms = (n/2) (a + l)
= (19/2) (105 + 195) = 2850
Hence the required sum is 2850.
Example 8: At what sum of all two-digit numbers, which, when divided by 4, yield unity as a remainder?
Solution:
The given numbers are 13, 17, ….. 97.
This is an AP with first term 13 and common difference 4.
Let the number of terms be n.
Then 97 = 13 + (n − 1)4
⇒ 4n = 88
⇒ n = 22
Therefore, the sum of the numbers = 222[13 + 97] = 11(110) = 1210.
Example 9: If a_{1}, a_{2}, a_{3},…, a_{n }are in A.P. with common difference , d, then the sum of the following series
Solution:
As given
Example 10: If a_{1}, a_{2}, a_{3},…, a_{n }are in A.P., where a_{i} > 0 for all i, then the value of
Solution:
As given
Example 11:
Solution:
The given numbers are in A.P.