Arithmetic Progression Solved Examples for JEE

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.

Arithmetic Progression Sequence

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_nis 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	\(\begin{array}{l}S_n = \frac{n}{2} [a + l]\end{array} \)

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^thterm is 3n and the second factors are 9, 12, 15, 18

t_n= [9 + (n − 1)3] = (3n + 6)

Hence n^thterm 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 a₁, a₂, a₃,…, a_nare in A.P. with common difference , d, then the sum of the following series

\(\begin{array}{l}\sin d(\cos \text{ec}\,{{a}_{1}}.co\text{sec}\,{{a}_{2}}+\text{cosec}\,{{a}_{2}}.\text{cosec}\,{{a}_{3}}+………..+\text{cosec}\ {{a}_{n-1}}\text{cosec}\ {{a}_{n}})\end{array} \)

Solution:

As given

\(\begin{array}{l}d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=….={{a}_{n}}-{{a}_{n-1}} \\\sin d\,\{co\text{sec}\ {{a}_{1}}co\text{sec}\ {{a}_{2}}+…..+\text{cosec}\ {{a}_{n-1}}\text{cosec}\ {{a}_{n}}\} \\=\frac{\sin ({{a}_{2}}-{{a}_{1}})}{\sin {{a}_{1}}.\ \sin {{a}_{2}}}+……+\frac{\sin ({{a}_{n}}-{{a}_{n-1}})}{\sin {{a}_{n-1}}\sin {{a}_{n}}} \\=(\cot {{a}_{1}}-\cot {{a}_{2}})+(\cot {{a}_{2}}-\cot {{a}_{3}})+….+(\cot {{a}_{n-1}}-\cot {{a}_{n}}) \\=\cot {{a}_{1}}-\cot {{a}_{n}}.\\\end{array} \)

Example 10: If a₁, a₂, a₃,…, a_nare in A.P., where a_i > 0 for all i, then the value of

\(\begin{array}{l}\frac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+ ……..+\frac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}=\end{array} \)

Solution:

As given

\(\begin{array}{l}{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=…….={{a}_{n}}-{{a}_{n-1}}=d\end{array} \)

where d is the common difference of the given A.P. Also

\(\begin{array}{l}{{a}_{n}}={{a}_{1}}+(n-1)d\end{array} \)

. Then by rationalising each term,

\(\begin{array}{l}\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{1}}}}+\frac{1}{\sqrt{{{a}_{3}}}+\sqrt{{{a}_{2}}}}+….+\frac{1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{n-1}}}}\\ =\frac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}}{{{a}_{2}}-{{a}_{1}}}+\frac{\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}}{{{a}_{3}}-{{a}_{2}}}+…..+\frac{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}}{{{a}_{n}}-{{a}_{n-1}}}\\=\frac{1}{d}\left\{ \sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}+\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}+……+\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}} \right\}\\=\frac{1}{d}\left\{ \sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}} \right\}=\frac{1}{d}\left( \frac{{{a}_{n}}-{{a}_{1}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}} \right)\\=\frac{1}{d}\left\{ \frac{(n-1)d}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}} \right\}\\=\frac{n-1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}.\end{array} \)

Example 11:

\(\begin{array}{l}\text{If}\ 1,\,\,{{\log }_{9}}({{3}^{1-x}}+2),\,\,{{\log }_{3}}({{4.3}^{x}}-1)\ \text{are in A.P.},\end{array} \)

then x equals

Solution:

The given numbers are in A.P.

\(\begin{array}{l}2{{\log }_{9}}({{3}^{1-x}}+2)={{\log }_{3}}({{4.3}^{x}}-1)+1\\ 2{{\log }_{{{3}^{2}}}}({{3}^{1-x}}+2)={{\log }_{3}}({{4.3}^{x}}-1)+{{\log }_{3}}3 \\ \frac{2}{2}{{\log }_{3}}({{3}^{1-x}}+2)={{\log }_{3}}[3({{4.3}^{x}}-1) \\ {{3}^{1-x}}+2=3\,({{4.3}^{x}}-1) \\ \frac{3}{y}+2=12y-3, \text{where}\ y={{3}^{x}} \\ 12{{y}^{2}}-5y-3=0 \\y=\frac{-1}{3} \text{or}\ \frac{3}{4}\Rightarrow {{3}^{x}}=\frac{-1}{3}\,\, \\\text{or}\ {{3}^{x}}=\frac{3}{4} x={{\log }_{3}}\,(3/4) \\\Rightarrow x=1-{{\log }_{3}}4.\end{array} \)