ARITHMETIC PROGRESSION

Question 1 : 

The first term of an A.P is 6 and the common difference is 5. Find the arithmetic progression its general term.

Solution :

a₁  =  6

d  =  5

Arithmetic Progression :

a₁, (a₁ + d), (a₁ + 2d), (a₁ + 3d),....................

Substitute 6 for a₁ and 5 for d. 

6, (6 + 5), (6 + 2 ⋅ 5), (6 + 3 ⋅ 5),....................

6, 11, 16, 21,....................

General Term : 

a_n  =  a₁ + (n - 1)d

Substitute 6 for a₁ and 5 for d. 

a_n  =  6 + (n - 1)5

a_n  =  6 + 5n - 5

a_n  =  5n + 1

Question 2 : 

Find the common difference and 15^th term of an arithmetic progression : 

125, 120, 115, 110,............... 

Solution :

a₁  =  125

a₂  =  120

Common Difference :

d  =  a₂ - a₁  

d  =  120 - 125

  d  =  -5

15^th Term :

a_n  =  a₁ + (n - 1)d

Substitute 15 for n, 125 for a₁ and -5 for d. 

a₁₅  =  125 + (15 - 1)(-5)

a₁₅  =  125 + (14)(-5)

a₁₅  =  125 - 70

a₁₅  =  55

So, the 15^th term of the given arithmetic progression is 55. 

Question 3 : 

Which term of the arithmetic progression 24, 23¼, 22½, 21¾, ……… is 3? 

Solution :

a₁  =  24

d  =  a₂ - a₁  =  23¼ - 24  =  -3/4

Let 3 be the nth term of the given arithmetic progression. 

Then, 

a_n  =  3

a₁ + (n - 1)d  =  3

Substitute 24 for a₁ and -3/4 for d. 

24 + (n - 1)(-3/4)  =  3

96/4 - 3n/4 + 3/4  =  3

(96 - 3n + 3) / 4  =  3

(-3n + 99) / 4  =  3

Multiply each side by 4. 

-3n + 99  =  12

Subtract 99 from each side. 

-3n  =  -87

Divide each side by -3.

n  =  29

Therefore, 3 is 29^th term in the given arithmetic progression.

Question 4 :

Find the 12th term of the arithmetic progression ;

√2, 3√2, 5√2,…………

Solution :

a₁  =  √2

a₂  =  3√2

Common Difference :

d  =  a₂ - a₁  

d  =  3√2 - √2

  d  =  2√2

12^th Term :

a_n  =  a₁ + (n - 1)d

Substitute 12 for n, √2 for a₁ and 2√2 for d. 

a₁₂  =  √2 + (12 - 1)(2√2)

a₁₂  =  √2 + (11)(2√2)

a₁₂  =  √2 + 22√2

a₁₂  =  23√2

So, the 12^th term of the given arithmetic progression is 23√2. 

Question 5 :

Find the 17th term of the arithmetic progression : 

4, 9, 14,…………

Solution :

a₁  =  4

a₂  =  9

Common Difference :

d  =  a₂ - a₁  

d  =  9 - 4

  d  =  5

17^th Term :

a_n  =  a₁ + (n - 1)d

Substitute 17 for n, 4 for a₁ and 5 for d. 

a₁₇  =  4 + (17 - 1)(5)

a₁₇  =  4 + (16)(5)

a₁₇  =  4 + 80

a₁₇  =  84

So, the 17^th term of the given arithmetic progression is 84. 

Question 6 : 

How many terms are in the following arithmetic sequence ?

7, 11, 15,……………483

Solution :

a₁  =  7

d  =  a₂ - a₁  =  11 - 7  =  4

Formula to find number of terms in an arithmetic progression :

n  =  [(l - a₁) / d] + 1

Substitute 7 for a₁ and 4 for d. 

n  =  [(483 - 7) / 4] + 1

n  =  [476 / 4] + 1

n  =  119 + 1

n  =  120

Therefore, there are 120 terms in the given arithmetic sequence. 

Question 7 : 

How many terms are there in the following Arithmetic progressions?

-16, -12, -8,……………16

Solution :

a₁  =  -16

d  =  a₂ - a₁

  d  =  -12 - (-16)

d  =  -12 + 16

d  =  4

Formula to find number of terms in an arithmetic progression :

n  =  [(l - a₁) / d] + 1

Substitute 16 for l, -16 for a₁ and 4 for d. 

n  =  [{16 - (-16)} / 4] + 1

n  =  [{16 + 16} / 4] + 1

n  =  [32 / 4] + 1

n  =  8 + 1

n  =  9

Therefore, there are 9 terms in the given arithmetic sequence. 

Question 8 : 

The 10^th and 18^th terms of an arithmetic sequence are 41 and 73 respectively. Find the 27th term

Solution :

Solving (1) and (2), 

a₁  =  5

d  =  4

27th Term : 

a₂₇  =  a₁ + (27 - 1)d

a₂₇  =  a₁ + 26d

Substitute 5 for a₁ and 4 for d. 

a₂₇  =  5 + 26(4)

a₂₇  =  5 + 104

a₂₇  =  109

So, the 27^th term of the arithmetic progression is 109. 

Question 9 :

Find n so that the n^th terms of the following two arithmetic progressions are equal. 

1, 7, 13, 19,................

and

100, 95, 90,................

Solution :

First Sequence : 

a₁  =  1

d  =  a₂ - a₁

  d  =  7 - 1 

d  =  6

Formula to find n^th term : 

a_n  =  a₁ + (n - 1)d

Substitute 1 for a₁ and 6 for d. 

a_n  =  1 + (n - 1)(6)

a_n  =  1 + 6n - 6

a_n  =  6n - 5

Second Sequence : 

a₁  =  100

d  =  a₂ - a₁

  d  =  95 - 100 

d  =  -5

Formula to find n^th term : 

a_n  =  a₁ + (n - 1)d

Substitute 100 for a₁ and 5 for d. 

a_n  =  100 + (n - 1)(-5)

a_n  =  100 - 5n + 5

a_n  =  105 - 5n

Given : n^th terms of the two arithmetic progressions are equal. 

Then,

6n - 5  =  105 - 5n

Add 5n and 5 to each side. 

11n  =  110

Divide each side by 11.

n  =  10

Question 10 :

How many two digit numbers are divisible by 13?

Solution :

Tow digit numbers :

10, 11, 12,………… 99

Find the two digit numbers divisible by 13. 

The first two digit number divisible by 13 is

13

The second two digit number divisible by 13 is

13 + 13  =  26

The third two digit number divisible by 13 is

26 + 13  =  39

................

The last two digit number divisible by 13 is 

91

Then, the two digit numbers divisible by 13 are 

13, 26, 39,.......................91

In the above sequence, the difference between any two consecutive term is 13. So, the above sequence is an arithmetic progression. 

Formula to find number of terms in an arithmetic progression :

n  =  [(l - a₁) / d] + 1

Substitute 91 for l, 13 for a₁ and 13 for d. 

n  =  [(91 - 13) / 13] + 1

n  =  [78 / 13] + 1

n  =  6 + 1

n  =  7

Therefore, there are seven two digit numbers divisible by 13.

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