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.

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 nth term of an AP Tn=a+(n1)dT_n = a + (n-1)d 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 Sn=n2[2a+(n1)d)]S_n = \frac{n}{2} [2a + (n-1)d)] 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 Sn=n2[a+l]S_n = \frac{n}{2} [a + l]

Also Read:

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 a1, a2,,an{{a}_{1}},\ {{a}_{2}},…………,{{a}_{n}}are in A.P. with common difference , d, then the sum of the following series sind(coseca1.coseca2+coseca2.coseca3+..+cosec an1cosec an)\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

d=a2a1=a3a2=.=anan1sind{cosec a1cosec a2+..+cosec an1cosec an}=sin(a2a1)sina1. sina2++sin(anan1)sinan1sinan=(cota1cota2)+(cota2cota3)+.+(cotan1cotan)=cota1cotan.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}}.\\

Example 10: If a1, a2, a3.an{{a}_{1}},\ {{a}_{2}},\ {{a}_{3}}…….{{a}_{n}} are in A.P., where ai>0{{a}_{i}}>0 for all i, then the value of 1a1+a2+1a2+a3+..+1an1+an=\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

a2a1=a3a2=.=anan1=d{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=…….={{a}_{n}}-{{a}_{n-1}}=d where d is the common difference of the given A.P. Also an=a1+(n1)d{{a}_{n}}={{a}_{1}}+(n-1)d. Then by rationalising each term,

1a2+a1+1a3+a2+.+1an+an1=a2a1a2a1+a3a2a3a2+..+anan1anan1=1d{a2a1+a3a2++anan1}=1d{ana1}=1d(ana1an+a1)=1d{(n1)dan+a1}=n1an+a1.\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}}}}.

Example 11: If 1,log9(31x+2),log3(4.3x1)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.

2log9(31x+2)=log3(4.3x1)+12log32(31x+2)=log3(4.3x1)+log3322log3(31x+2)=log3[3(4.3x1)31x+2=3(4.3x1)3y+2=12y3, where y=3x12y25y3=0y=13 or343x=13 or 3x=34x=log3(3/4)x=1log34.2{{\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.

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