Arithmetic Progression Solved Examples

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

Also Read:

• Arithmetic Progression
• Arithmetic Mean

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 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 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 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 ${{a}_{1}},\ {{a}_{2}},…………,{{a}_{n}}$are in A.P. with common difference , d, then the sum of the following series $\sin d(\cos \text{ec}\,{{a}_{1}}.co\text{sec}\,{{a}_{2}}+\text{cosec}\,{{a}_{2}}.\text{cosec}\,{{a}_{3}}+………..+\text{cosec}\ {{a}_{n-1}}\text{cosec}\ {{a}_{n}})$ is

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 $\frac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+ ……..+\frac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}=$

Solution:

As given

Example 11: If $1,\,\,{{\log }_{9}}({{3}^{1-x}}+2),\,\,{{\log }_{3}}({{4.3}^{x}}-1)$are in A.P. then x equals

Solution:

The given numbers are in A.P.