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nth Term of an AP


The nth term of an arithmetic progression whose first term is a1 and whose common difference is d is given by ana1 + (n – 1) d.

This formula tell us that if we know the first term

 a1 and the common difference d of an arithmetic progression, then we can arrive at any  term we  want.

Very Short Answer Type

 

Example

Find d and write the next four terms of each of the following arithmetic progression:
(i) 7, 11, 15, 19, 23, … 

Solution

Here a1 = 7, and d = 11 – 7 = 4.

We are given the first five terms of the arithmetic progression. The next four terms are given by

a6 = a5 + 4 = 23 + 4 = 27,

a7 = a6 + 4 = 27 + 4 = 31,

a8 = a7 + 4 = 31 + 4 = 35,

and a9 = a8 + 4 = 35 + 4 = 39.


 

Example

(ii) Find d and write the next four terms of each of the following arithmetic progression:

 

Solution

(ii) Here

a1 = 

We are given the first six terms of the arithmetic progression. The next four terms are given by


 

Example

 Find d and write the next four terms of each of the following arithmetic progression:

 a + 2ba + baa – ba – 2 b, …

Solution

Here a1 = a + 2 b and d = (a + b) – (a + 2 b) = − b.

We are given the first five terms of the arithmetic progression. The next four terms are given by

a6 = a5 – b = a – 2 b – b = a – 3 b,

a7 = a6 – b = a – 3 b – b = a – 4 b,

a8 = a7 – b = a – 4 b – b = a – 5 b,

and a9 = a8 – b = a – 5 b – b = a – 6 b.

a9 = a8 – b = a – 5 b – b = a – 6 b.
 


 

Example

 Find the 18th, 23rd and

nth terms of the arithmetic progression − 11, − 9, − 7, − 5, …

Solution

Here a1 = − 11, d = − 9 – (− 11) = 2. We have


a1 = − 11, d = − 9 – (− 11) = 2. We have

a18 = a1 + (18 – 1) d 

     = a1 + (17) d 
     = − 11 + (17) (2) 
     = − 11 + 34 
     = 23.

a23 = a1 + (23 – 1) d 
     = 
a1 + (22) d 
     = − 11 + (22) (2) 
     = − 11 + 44
     = 33.

and an = a1 + (n – 1) d 
          = − 11 + (
n – 1) (2)
          = − 11 + 2 
n – 2
          = 2
n – 13.


 

Example

Find the 25th, 30th and nth terms of the arithmetic progression π + 3, π + 1, π − 1, π − 3, …

Solution

(ii) Here a1 = π + 3
            
d  = (π + 1) – (π + 3) 
                = − 2. 

We have a25 = a1 + (25 – 1) d 
                  = a1 + (24) d 
= π + 3 + (24) (− 2) 
                  = π + 3 –48 
                  = π − 45.

a30 = a1 + (30 – 1) d 
     = 
a1 + (29) d
     = π + 3 + (29) (− 2) 
= π + 3 –58 
= π − 55.

and an = a1 + (n – 1) d 
          = π + 3 + (
n – 1) (− 2)
          = π + 3 – 2
n + 2
= π + 5 – 2n

 

 
Example

Find the 9th, 16th and nth terms of the arithmetic progression a – 3ba + ba + 5 ba + 9 b, …   

Solution

Here a1 = a – 3 b

a1 = a – 3 b

              d  = a + b – (a – 3 b
                  = 
a + b – a + 3 b 
= 4 b

We have a9 = a1 + (9 – 1) d 
                 = (
a – 3 b) + 8(4 b
                 = 
a – 3 b + 32 b 
                 = 
a + 29 b.

a16 = a1 + (16 – 1) d 
     = (
a – 3 b) + (15) (4 b
     = 
a – 3 b + 60 b 
     = 
a + 57 b

and an = a1 + (n – 1) d 
          = (a – 3 b) + (n – 1) (4 b)
          = 
a – 3 b + 4b – 4 b 
          = 
a + (4n – 7) b.

 

 

Example

Find the 20th, 40th and nth terms of the arithmetic progression 

Solution


 

Example

(i) Find the indicated term(s) in each of the following arithmetic progressions:a1 = 3, d = − 2; a10an 

Solution

We havea1 = 3, d = − 2. Using the formula an = a1 +(n – 1d we get,

a1 = 3, d = − 2. Using the formula an = a1 +(n – 1d we get,

an = 3 + (n – 1) (− 2) = 3 – 2 n + 2 = 5 – 2 n.

In particular, a10 = 5 – 2(10) = 5 – 20 = − 15.

(ii) Find the indicated term(s) in each of the following arithmetic progressions:
a1 = − 1, d = 2; a15an
We have a1 = − 1, d = 2. Using the formula an = a1 + (n – 1d we get,

an = (− 1) + (n – 1) (2) = − 1 + 2 n – 2 = 2 n – 3.

In particular, a15 = 2 (15) –3 = 30 – 3 = 27.

 

 
Example

 Find the indicated term(s) in each of the following arithmetic progressions:

 a1 = 

Solution

We have a1 = . Using ar = a1 + (– 1) d, we get

ar = 

How to Shorten your Calculation?
If we know one term of an arithmetic progression, say ar, and the common difference d of the arithmetic progression, we can write any term of the arithmetic progression using ar and d.

We have an = a+ (n – 1) d 

                 =

a1 + (r – 1) d + (n – rd 

                 = ar + (n – rd.
 
 

 

Example

Determine the 25th term of the arithmetic progression whose 9th term is –6 and common difference is –.

Solution

We are given that

a9 = − 6  and d = − .

We know that a9 = a1 + (9 – 1) d = a1 + 8 d.

Now, a25 = a1 + (25 – 1) d = a1 + 24 d

             = (a1 + 8 d) + 16 d = − 6 + 16 (− 3/2) = − 6 – 24 = − 30.


 

Example

The first term of an arithmetic progression is – 2 and its 10th term is 16. Determine its 15th and 20th terms.

Solution

We have

a1 = − 2 and a10 = 16.

We know that

a10 = a1 + (10 – 1) d = a1 + 9 d.

16 = − 2 + 9 d  18 = 9 d  d = 2.

Now,

a15 = a1 + (15 – 1) d = − 2 + (14) (2) = − 2 + 28 = 26,

and

a20 = a1 + (20 – 1) d = − 2 + (19) (2) = − 2 + 38 = 36.
 


Example

Ifpth term of an arithmetic progression is q and its qth term is p, show that its rth term is p + q – r. What is its (p + q)th term?

Solution

(iii) We get given that

ap = q and aq = p.

This implies

a1 + (p –1d = q.... (1) and a1 + (q – 1d = p....(2)

a1 – d + p d = q and a1 – d + q d = p

Subtracting the second equation from the first, we get

p d – q d = q – p

or

(p – qd = − (p – q) or d = − 1.

Substituting

d = − 1 in (1) we get,

a1 + (p –1) d q.... (1)

a1 + (p –1) (− 1) = q

aq + p − 1

Now ar = a1 + (r – 1) d

           =

a1 + (p – 1) d +  (r– 1) d − (p – 1) d       (Adding and subtracting (p – 1) d )

           =

q + (r – 1)( − 1) − (p – 1) (− 1)                    (Substituting (1) and d = − 1) 

           =

q − r + 1 + p − 1 

           =

q + p − r

Thus

arq + p − r

But,

ap aq  = a p+q = a1 + (p – 1d + a1 + (q – 1d = 2 a+( p − q − 1) d

                             = 2 a1+( p + q − 2) d.... (3)

Substituting

aq + p  1, d = − 1 in (3) we get

Thus,

ap + q = 2(q + p − 1) + (p + q − 2)( − 1)

               = 2

q + 2 p − 2 − p − q + 2

               =

p + q


 

Example

The fourth term of an arithmetic progression is equal to 3 times the first term and the seventh term exceeds twice the third term by 1. Find its first term and the common difference. 

Solution

(iv) We are given that

a4 = 3 a1 and a7 = 2 a+ 1. 

This implies 
a1 + 3 d = 3 a1 and a1 + 6 d = 2 (a1 + 2 d) + 1  

 3 d = 2 a1 and 2 d = a1 + 1.

Putting

a1 =  d in 2 d = a1 + 1, we get 2 d =  d + 1 
or 2
 d -  d = 1

or 

d = 1 or d = 2.

Also

a1 =  d = 

Hence a1 = 3 and d = 2.

 

 

 

Example

If (p + 1)th term of an arithmetic progression is twice the (q + 1)th term, show that (3 p + 1)th term is twice the (p + q + 1)th term.

Solution

(v) We are given that 

[a1 + {(p +1) – 1} d] = 2 [a1 + {( q + 1) – 1} d]

a1 + p d = 2 (a1 + q d).                                                     (1)

Now (3

p + 1)th term = a1 + (3 p + 1 – 1) d = a1 + 3 p d          (2)

Thus, adding 2

p on both sides of (1), we get

a1 + p d + 2 p d = 2 (a1 + q d) + 2 p d

a1 + 3 p d = 2 (a1 + q d + p d) = 2 [a1 + {( p + q +1) – 1} d]

Here, LHS is nothing but the (3

p + 1)th term as seen in (2) and RHS is twice the (p + q + 1)th term. Hence proved.
 

 

 

Example

If 7 times the 7th term of an arithmetic progression is equal to 11 times its 11th term, show that the 18th term of the arithmetic progression in zero.

Solution

We are given that 7

a7 = 11 a11  

                    

 7[a1 + (7-1) d] = 11 [a1 + (11-1) d]
                            7[a1 + 6 d] = 11 [a1 + 10 d

                           

 7 a1 + 42 d = 11 a1 + 110 d
                                         0 = 4 a1 + 68 d 

                         

 4[a1 + 17 d] = 0

                           

 a1 + 17 d = 0.

Note that the left hand side is nothing but 
a18. Thus, a18 = 0.


 

Example

If m times the mth term of an arithmetic progression is equal to n times its nth term, show that (m + n)th term of the arithmetic progression is zero. [Assume that m  n.]

Solution

We are given that

m am = n an

m [a1 + (m - 1)d] = n [a1 + (n - 1) d]

m [a1 + m d – d] = n [a1 + n d – d]

m [α + m d] = n [α + n d] where α = a1 – d

m α + m 2 d = n α + n 2 d 

(m – nα + (m 2 – n 2 ) d = 0

(m – n) [α + (m + n) d] = 0 

α + (m + nd = 0 (or) m = n but m = n is not possible as m  n (given)

a1 + (m + n – 1) d = 0             [Q α = a1 – d]

Note that left hand side is nothing but

m + n. This m + n = 0

 

Example

 If mth term of an arithmetic progression is and its nth term is. Find the (m n)th term of the arithmetic progression. [Assume that m  n.]

Solution

We are given that

am =  and an =.

a1 + (m – 1) d = and an = (n – 1) d = 

Subtracting the second equation from the first, we get

Now, 

a m n  = a1 + (m n – 1) 

                 =

a1 + m n d – d = a1 – d + m n d

                 =

a1 + m d – d – m d + m n d     [adding and subtracting m d]

=

a1 +( m – 1) d – m d + m n d     

                          

 


 

Example

Which term of the arithmetic progression 

Solution

We are given that .

Suppose −  is the

nth term of the given arithmetic progression, then

Thus, −  is the 11th term of the arithmetic progression.

 


 

Example

Is  ( 31) a term of the arithmetic progression  1, 5, 11, 17, …?

Solution

(ii) We are given that

a1 = − 1, d = 5 − (− 1) = 6.

Let − 31 be the

nth term of the above AP

   − 31 =

an  

         =

a1 + (n – 1d 

         = − 1 + (

n – 1)(6)

⇒ −

30 = 6n – 6 

− 24 = 6n 

  

 n = − 4

Since

n is not a positive integer, − 31 cannot be a term of the given arithmetic progression.
 


 

Example

 Which term of the arithmetic progression

 

Solution

We are given that

Suppose  is

nth term of the given arithmetic progression.

Then 

11p + 24q+ 3 p +4 q = (p + 2 q) (n− 1)

        [cancel –from both the sides]

14 p + 28 q = (p + 2 q) (n − 1)

14(p + 2 q) = (p + 2 q) (n − 1)

          

 14 = n – 1

          

 n = 15.        [cancel p + 2 q from both the sides]

Thus,  is 15th term of the given arithmetic progression.

 


 

Example

Which term of the arithmetic progression

Solution

We are given that 

Suppose is the

r th term of the given arithmetic progression, then

Canceling  from both the sides, we get

n – 5 = 3 n – 1 – 2r + 2 = 3 n + 1 – 2 r

2 r = 3 n + 1 – n + 5 = 2 n + 6  r = n + 3.

Thus,  is (

n + 3)th term of the given arithmetic progression.

 


Effect of Addition and Multiplication on an AP

 

If we add (subtract) the same constant k from each term of an AP, the resulting sequence is still an AP.

Let the given AP be

a, a + d, a + 2 d, a + 3 d, a + 4 d, …                     (1)

The nth term of this AP is

a n = a + (n – 1) d.


The sequence obtained by adding a constant

k to each term of (1) is

a + k, a + d + k, a + 2 d + k, a + 3 d + k, …

or

a + k, (a + k) + d, (a + k) + 2 d, (a + 3 d) + k, …


This is an AP with first term

a + k and common difference d.


Also, note that if we multiply (divide) each term of an AP by the same constant (non-zero constant) then the resulting sequence is an AP

If we multiply each term of (1) by

k, we obtain the sequence

ka, k a + k d, k a + 2k d, k a + 3k d, …


This is an AP with first term

ka and common difference k d.

Points to Note in an A.P.

In view of the above discussion, to prove

a, b, c are in AP is equivalent to prove any one of the following


(i) a + k, b + k, c + k are in AP          
(ii)
ak, bk, ck are in AP

(iii) k a, k b, k c are in AP (k 0)     
(iv) are in AP (
k 0)
 

We can also say that addition, subtraction, multiplication and division of each term by the constant does not affect the nature of an AP.
 

Of course we should not forget that if

abc are in AP.

⇔ b − a = c − b 2 b = a + c.

 

Example

 If abc are in AP, Show that

 (i) are in AP    (assume abc ≠ 0).

 (ii) b + cc + aa + b are in AP.

 (iii) are in AP.

 (iv) (b c)2− a2, (c + a)2− b2, (a + b)2− c2 are in AP.

           (assume that a + b + c  0) [ 1985]

Solution

(i) abc are in are in AP

[division of each term by a non-zero constant does not affect the nature of the AP]

are in AP

Alternative Solution

abc are in AP

     [divide by

abc]


(ii) abc are in AP


⇒  a− b− c are in AP   [multiplication by the same constant (− 1)]

⇒ a + bc − aa + bc − bab + c − c are in AP  

                              [addition of the same constant (a + b + c)]

⇒ b + cc + aa + b are in AP

Alternative Solution

b + cc + aa + b are in AP

( c + a) – (b + c) = (a + b) - (c + a

a − b = b − c 

b – a = c – b   [multiply by – 1]

abc are in AP

Since the last statement is a true statement, we get the first statement, viz.

b + cc + aa + b are in AP is a true statement.


(iii) are in AP

(since abc are in AP b − a = c − b)

⇔ abc are in AP.


(iv) abc are in AP

⇒ − 2a, − 2b, − 2c are in AP. [multiplication by the same constant (− 2)]

a + b + c − 2aa + b + c – 2ba + bc – 2c are in AP

                                                                          [addition of the same constant (a + b + c)]

b + c − ac + a – bab – c are in AP.

(a + b + c) (b + c – a), (a + b + c) (c + a – b), (a + b + c) (a + b – c) are in AP.

                                    [multiplication by the same constant (ab + c)]

(b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c 2 are in AP.

 

 

Example

Determine k so that k + 2, 4 k – 6 and 3 k – 2 are three consecutive terms of an A.P.

Solution

(i) k + 2, 4 k – 6, 3 k – 2 are in AP

⇔ 2(4 k – 6) = (k + 2) + (3 k – 2)  

⇔ 8 k – 12 = 4 k  
⇔ 4 k = 12
⇔ k = 3.

Thus, the required value ofk = 3.

 

 

Example

Determine k so that kk 2 + 5 k + 2, 3 k are three consecutive terms of an A.P.

Solution

kk 2 + 5 k + 2, 3 k are in AP

⇔ 2(k 2 + 5 k + 2) = k + 3 k 

⇔ k 2 + 5 k + 2 = 2 k 
⇔ k 2 + 3 k + 2 = 0

⇔ (k + 1) (k + 2) = 0

⇔ k = − 1, − 2.


 

Example

If a2b2c2 are in AP, show that are in AP.

Solution

 

b2− a2 = c2− b2⇔ a2b2c 2 are in AP.

 

 

Example

If a2b2c2 are in AP, show that are in AP.

Solution

    

⇔ (b- a) (b + a) = (c - b) (c + b⇔ a2b2c2 are in AP.


 

Example

  If a2 (b + c), b2 (c + a), c2 (a + b) are in AP. 

and b c + c a + a b  0, show that abc are in AP.

Solution

a2(b+ c), b2(c + a), c2(a +b) are in AP.

⇔ b2(c + a) – a2 (b + c) = c2 (a + b) – b2 (c + a)

⇔ b2 c + b2 a – a2 b – a2 c = c2 a + c2 b – b2 c – b2 a

⇔ c (b2 a2) + b a (b a) = a (c2 – b2) + b c (cI> - b)

⇔ (b- a) { c (b+ a) + b a } = (c -b) { a ( c +b) + b c } (canceling common term (a b + b c + c a) on both sides)

⇔ b – a = c – b ⇔ abc are in AP.              [b c + c a + a b  0]

 

 

Example

If are in AP, show that  are in AP.

Solution

(vi) are in AP

 are in AP
                                            [adding the same constant "2"]

are in AP

are in AP. [Dividing by the same constant "(

a + b + c)"]


 

Example

If ab and c be respectively the pthqth and rth terms of an AP, show that a (q – r) + b (r - p) + c (p – q) = 0.

Solution

(i) Let α denote the first term and β the common difference of the AP, of which ab, and c, are pthqth and rth terms.

a = α + (p - 1) β = α + pβ - β                    (1)

b = α + (q - 1) β = α + qβ - β                    (2)

c α + (r - 1) β = α + rβ - β                      (3)

Note that we have to eliminate α and β from these three equations.

Subtracting (2) from (1) and (3) from (2), we get

(a – b) = (p – qβ                                    (4)

and (b – c) = (q – rβ                              (5)

Dividing (4) by (5), we get

⇒ ( a – b) (q – r) – (b - c) (p - q) = 0

⇒ a ( q - r) - b (q - r + p - q) + c (p - q) = 0

⇒ a ( q - r) + b (r - p) + c (p - q) = 0.

 

 

Example

If ab and c are three terms of an AP, such that a  b, show that is a rational number.

Solution

Let ab and c be the pthqth and rth terms of the AP, and let a1 be its first term and d the common difference. 

Then a a1 + (p-1) db = a1 + (q –1) d and c a1 + (r-1) d
Since a  0 and d  0, we can write

,

which is a rational number, since pq and r are integers. 


 





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