CBSE Tutorials

nth Term of an Arithmetic Progression

    Let …….. be an A.P., with first term as a, and common difference as d.

    First term is         = a            (i)

    Second term     = a + d            (ii)

                =

    Third term         =         (iii)

                =      [from (i)]

                =

        or         =

    Fourth term

        or          =      [from (iii)]

                 =

                 =

    \      nth term a_n

Note: The nth term of the A.P. with first term a & common difference d is given by

is also called as general term of an A.P.

If there are P terms in the A.P. then a_p represents the last term which can also be denoted by l.

Problems based on nth TERM OF AN Arithmetic Progression

Problem:

Find the 18^th term and n^th term for the sequence 7, 4, 1, –2, –5.

Solution:

Here      a = 7

        and      d =

             = 4 – 7 = – 3

             n = 18

        

        

             = 7 + 17 ´ –3

             = 7 – 51

             = –44

            

             = 7 + (n – 1) (–3)

             = 7 – 3n + 3

             = 10 – 3n

Problem:

Which term of the A.P. 7, 12, 17, ….. is 87?

Solution:

a = 7

             

              = 12 – 7 = 5

            

        As     

            87 = 7 + (n – 1) ´ 5

        

        

            

            17 = n

        \     17^th term of give A.P. is 87

Problem:

How many terms are there in A.P. 7, 13, 19, …….., 205.

Solution:

a = 7

        

         = 13 – 7 = 6

        

        

        205 – 7 = 6n– 6

        198 + 6 = 6n

         or 34 = n

        \     Given A.P. has 34 terms

Problem:

Check whether –150 is term of the A.P. 11, 8, 5, 2, …….?

Solution:

a = 11

        

         = 8 – 11 = -3

        let                 [Assume]

        

        

        

        

        As number of term can’t be in fraction, \ –150 is not a term of the given A.P.

Problem:

In the following A.P. find the missing terms in the boxes.

        (i) 5,  ,  ,             (ii) –4,  ,  ,  ,  , 6

Solution:

(i)    Here a = 5

            & or

        Now,     

            

            

            

            

        \    

        \    

            

        Hence     

        (ii)     Here

            

        As    

             6 = –4 + 5d

            

             d = 2

        \     

            

            

            

        

Problem:

For what value of n, the nth terms of A.P’s 63, 65, 67, ……… and 3, 10, 17, …… are equal.

Solution:

First sequence is 63, 65, 67, ……..

            a = 63

            d₁ = 65 – 63 = 2

            

             = 63 + 2n–2

             = 61 + 2n

        Second sequence is 3, 10, 17, ………..

        \

        \

         =    

         = 7n–4

        According to question

        61 + 2n = 7n– 4

        61 + 4 = 7n– 2n

          65 = 5n

        

        \n = 13

Problem:

Two A.P.s have the same common difference. If the difference between their 100^th terms is 100, what is the difference between their 1000^th terms?

Solution:

Let the common difference of two A.P.s be d

        Then, their 100^th terms will be

            

        According to question     

        i.e.            …(i)

        Now, difference between their 1000^th terms

            

                    [By equation (i)]

Problem:

Which term of the arithmetic progression 3, 10, 17, …. will be 84 more than its 13^th term?

Solution:

The give A.P. is 3, 10, 17, ……

        Here, a (first term)     = 3

        d (common difference)= 10 – 3 = 7

                

                

                     = 3 + 12 ´ 7 = 3 + 84 = 87

        Let n^th term be 84 more than the 13^th term of the given A.P.

        So, we get    

            

            

            

        

        

        \ The 25^th term of the given A.P. will be 84 more than its 13^th term.

Problem:

Find the 31^st term of an A.P. whose 11^th term is 38 and the 16^th term is 73.

Solution:

Let a be the 1^st term and d the common difference.

        Here                    …(i)

                            …(ii)

        Subtracting (ii) and (i), we get

            

    or            

        Putting d = 7 in (i), we get

            

            

        \     

        Hence, 31^st term is 178.

Problem:

How many three digit numbers are divisible by 7?

Solution:

Three digit numbers which are divisible by 7 are 105, 112, 119, …, 994.

        Here,     

        \    

            

        Þ    

        Þ        

        Þ         n – 1 = 127

                 n = 127 + 1 = 128.

        \    There are 128 three digit numbers which are divisible by 7

Problem:

A sum of Rs 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests from an AP? If so, find the interest at the end of 30 years.

Solution:

We know that the formula to calculate simple interest is given by

            Simple Interest =

        So, the interest at the end of the 1^st year     = Rs = Rs 80

        The interest at the end of the 2^nd year     = Rs = Rs 160

        The interest at the end of the 3^rd year     = Rs = Rs 240

        Similarly, we can obtain the interest at the end of the 4^th year, 5^th year, and so on.

        So, the interest (in Rs) at the end of the 1^st, 2^nd, 3^rd, ….. years, respectively are 80, 160, 240, …

        It is an AP as the difference between the consecutive terns in the list is 80, i.e., d = 80. Also, a = 80.

        So, to find the interest at the end of 30 years, we shall find .

        Now,

        So, the interest at the end of 30 years will be Rs 2400.

TO FIND nth TERM FROM THE END OF AN A.P.

    Consider the following A.P. ,

    where l is the last term

    last term          = – (1 – 1)d

    2^nd last term     –
d = – (2 – 1) d

    3^rd last term      – 2d = – (3 – 1)d

            ……………………………

            ……………………………

Note: nth term from the end =  – (n – 1) d

Problem:

Find the 5^th term from the end of the AP, 17, 14, 11, ….., –40

Solution:

1^st method

        

        Using

        5^th term from the end will be

        

        

        

        = –28

2^nd method

        Sequence can be written as –40, –37, ….11, 14, 17

        \     

            

             =–37 + 40

             = 3

         n = 5

        Using

             =

            

             = –40 + 12

             = –28

Problem:

The sum of the 4^th and 8^th terms of an A.P. is 24 and sum of 6^th & 10^th terms is 44. Find the first three term of the AP.

Solution:

Using 1^st condition

        \     

            

            

        or                 …(i)

        Using 2^nd condition

            

        or    

            

        or                …(ii)

        Subtracting equation (i) from (ii) we have

            

        or     d = 5

        Putting value of d in equations (i)

            

            

            

        \

            

        

            

            

Problem:

If the pth term of an A.P. is q and the qth term is p, prove that its nth term is

Solution:

Let a be the first term and d be the common difference of the given A.P.

        Then,     pth term = q    Þ        …(i)

            qth term = p    Þ        …(ii)

        Subtracting equation (ii) from equation (i), we get

                    …(iii)

        Putting in equation (i), we get

            

        

        \    nth term

                 [from equation (iii)]

            

Problem:

If m times the mth term of an A.P. is equal to n times its nth term, show that the (m + n) term of the A.P. is zero.

Solution:

Let a be the first term and d be the common difference of the given A.P.

        According to question, m times the mth term = n times the nth term

            Þ    

            Þ    

            Þ    

            Þ    

            Þ    

            Þ    

            Þ    

            Þ    

             Þ            []

            Þ    

        \     th term of A.P. is zero

CONDITION FOR TERMS TO BE IN A.P.

    If three numbers a, b, c, in order are in A.P. Then,

         common difference =

    Þ    

    Þ    

Note: a, b, c are in A.P. iff

Problem:

If are in A.P., find the value of x.

Solution:

Since, are in A.P.

        \     

        Þ    

        Þ    

        Þ     x = 6

Problem:

If the numbers a, b, c, d, e form an A.P., then find the value of .

Solution:

Let D be the common difference of the given A.P. Then,

             and

        \    

        Þ    

        Þ    

        Þ