Arithmetic Progression

Arithmetic Progression

Sequence

A sequence is a set of numbers written in a particular order.

E.g.         1, 3, 5, 7, 9, … is a set of odd numbers.

                1, 4, 9, 16, … is a set of square numbers.

Series

A series is a sum of terms of a sequence. If there are n terms in the sequence and we evaluate the sum then we often write,

                S_n = u₁ + u₂ + u₃ + ……………. + u_n

Arithmetic Progression

Let’s consider two common sequences, 1, 3, 5, 7, … and 0, 10, 20, 30, …

It is easy to see how these sequences are formed. They start with a particular first term and then to get successive terms we just add a fixed value to the previous term. In the first sequence, we add 2 to get next term where as in the second sequence we add 10. So, the difference between consecutive term in each sequence is constant.  We could also subtract a constant. eg. in a sequence 8, 5, 2, -1, -4, … the difference between the consecutive terms is -3. Any sequence with this property is called arithmetic progression or A.P. for short.

If ‘a’ stands for the first term of a sequence and ‘d’ stands for the common difference between the successive terms then,

First term (t₁)                     = a

Second term (t₂)                               = a + d

Third term (t₃)                   = a + d + d = a + 2d

Fourth term (t₄)                = a + 3d

. . .

n^th term (t_n)                        = a + (n – 1) d   

Hence, t_n = a + (n – 1) d

Sometimes, we also write ‘l’ for the last term of a finite sequence. So, we also write

l = a + (n – 1) d

1.  Write down the first five terms of A.P. with first term 8 and common difference 7.

Solution,

First term (t₁)     = a          = 8

Common difference (d) = 7

So, second term (t₂)        = a + d

                                                = 8 + 7 = 15

Third term (t₃)                   = a + 2d

                                                = 8 + 2 × 7 = 22

Fourth term (t₄)                = a + 3d

                                                = 8 + 3 × 7 = 29

Fifth term (t₅)                     = a + 4d

                                                = 8 + 4 × 7 = 36

Hence, the first five terms of an A. P. with first term 8 and common difference 7 are 8, 15, 22, 29 and 36.

2.  Find the eighth term of the sequence 7, 11, 15, 19, ……

Solution,

First term (a) = 7

Common difference (d) = 11 – 7 = 4

Now, eighth term (t₈)     = a + 7d

                                                = 7 + 7 × 4 = 35

3.  If the 4^th term and the common difference of an A.P. are 27 and 5 respectively, find the 1^st term.

Solution,

Here,     Common difference (d) = 5         

                4^th term (t₄)         = 27

or,          a + 3d = 27

or,          a + 3 × 5 = 27

or,          a = 27 – 15

                a = 12

4.  If there be 60 terms in A.P. of whose the first term is 8 and last term is 185, find the 21^st term.

Solution,

Here,     last term (t_n) = 185

or,          a + (n – 1) d = 185

or,          8 + (60 – 1) d = 185

or,          59d = 185 – 8

or,          d =  = 3

Now, 21^st term (t₂₁) = a + 20d

                                = 8 + 20 × 3

                                = 68

5.  How many terms are there in the series 1 + 4 + 7 + …… + 64?

Solution,

First term (a) = 1

Common difference (d) = 4 – 1 = 3

Here,     last term (t_n)       = 64

or,          a + (n – 1) d = 64

or,          1 + (n – 1) × 3 = 64

or,          1 + 3n – 3 = 64

or,          3n = 64 + 2

or,          n =  = 22

hence, there are 22 terms in the given series.

.  An A.P. is given by k, ,  , 0, …….

a.  Find the sixth term.

b.  Find the n^th term.

c.  If the 20^th term is 15, find k.

Solution,

Here,     the given sequence is k, 2k/3, k/3, 0, …….

So,          common difference (d) = t₂ – t₁

                                                                = 2k/3 – k = (2k – 3k)/3 = (-k)/3

a.  Now,                sixth term (t₆)    = a + 5d

                                                = k + 5 × (-k)/3

                                                = k – 5k/3                             = (-2k)/3

b.            n^th term (t_n)         = a + (n – 1) d

                                                = k + (n – 1) (-k)/3

                                                = k – (kn)/3 +k/3              

                                                = (4k – kn)/3

                                                = {k (4 – n)}/3

c.            20^th term (t₂₀)    = 15

                a + 19d                  = 15

                k + 19 × (-k)/3    = 15

                (-16k)/3                = 15

                k                              = (-15 x 3)/16    

                                                = - 45/16

7.  If the 5^th term of an A.P. is 19 and the 8^th term is 31, which term is 67?

Solution,

Since,    t_n = a + (n – 1) d

                5^th term (t₅)         = 19

or,          a + 4d    = 19

                a              = 19 – 4d -------------(i)

                8^th term (t₈)         = 31

or,          a + 7d    = 31

or,          a              = 31 – 7d -------------(ii)

From equation (i) and (ii),

19 – 4d = 31 – 7d

or,          7d – 4d = 31 – 19

or,          3d           = 12

or,          d             = 12/3   = 4

putting the value of d in equation (i),

                a              = 19 – 4 × 4

or,          a              = 3

Let,        n^th term                = 67

Then,     a + (n – 1) d         = 67

or,          3 + (n – 1) 4        = 67

or,          4n – 1                    = 67

or,          n                             = 68/4   = 17

Hence, 67 is the 17^th term.  

8.  The 5^th term of an A.P. is 28 and the 9^th term is 48. Are 63 and 95 terms if this A.P.?

Here,     5^th term = 28

i.e.          a + 4d = 28

a = 28 – 4d ------------------(i)

again,    9^th term = 48

i.e.          a + 8d = 48

                a = 48 – 8d ------------------(ii)

from (i) and (ii),

28 – 4d = 48 – 8d

or,          8d – 4d = 48 – 28

                d = 20/4                = 5

putting the value of d in (i),

                a = 28 – 20

                a = 8

if possible, let 63 is the n^th term of A.P.

then,     a + (n – 1) d = 63

or,          8 + (n – 1) 5 = 63

 or,         5n + 3 = 63

or,          5n = 60

or,          n = 60/5                = 12

therefore, 63 is the 12^th term of A.P.

if possible, let 95 is the x^th term of A.P.

then,     a + (x – 1) d = 95

or,          8 + (x – 1) 5 = 95

or,          5x + 3 = 95

or,          5x = 92

or,          x = 92/5

therefore, 95 is not any term in A.P. because x is fraction, which is not possible.

9.  The n^th term of series 9 + 7 + 5 + … is same as the n^th term of the series 15 + 12 + 9 + … Find n.

Solution,

For the series 9 + 7 + 5 + …

First term (a) = 9

Common difference (d) = 7 – 9 = - 2

n^th term (t_n)         = a + (n – 1) d     = 9 + (n – 1) × (-2)            

                                                                = 11 – 2n

For the series 15 + 12 + 9 + …

First term (a) = 15

Common difference (d) = 12 – 15 = - 3

n^th term (t_n)        = a + (n – 1) d     = 15 + (n – 1) × (-3)

                                                                = 18 – 3n

By question,

11 – 2n = 18 – 3n

or,          3n – 2n = 18 – 11

or,          n             = 7