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 nth term of an AP | Where d = an – an-1 and Tn is the nth term of the sequence. | |
The formula for finding the sum to n terms of an AP | where Sn 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 |
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Solved Examples on Arithmetic Progression
Example 1: Let the bth term of an A.P. be q and cth term be p, then find its dth term.
Solution:
Tb = a + (b − 1)d = q …..(i) and
Tc = a + (c − 1)d = p …… (ii)
From (i) and (ii),
we get d = −(b−c)/(b−c) = −1
Putting value of d in equation (i), then a=b+c−1
Now, dth term is given by A.P.
Td = 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,
Sn = 112/2 (104 + 992) = 61376
Example 3: If the 8th term of an AP is zero, then what is the ratio of its 28th and 18th term?
Solution:
Given that 8th term = a+(8−1)d = 0
⇒ a+7d = 0
Now ratio of 28th and 18th 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 nth 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 nth term is 3n and second factors are 9, 12, 15, 18
tn=[9+(n−1)3]=(3n+6)
Hence nth 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 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
Solution:
As given
Example 10: If
Solution:
As given
Example 11: If
Solution:
The given numbers are in A.P.