CBSE 10th Mathematics | Arithmetic Progressions | nth Term of an Arithmetic Progression

nth Term of an Arithmetic Progression

    Let \displaystyle {{a}_{1}},{{a}_{2}},{{a}_{3}}…….. be an A.P., with first term as a, and common difference as d.

    First term is         = a            (i)

    Second term \displaystyle ({{a}_{2}})     = a + d            (ii)

                = \displaystyle a+(2-1)d

    Third term \displaystyle ({{a}_{3}})        = \displaystyle {{a}_{2}}+d        (iii)

                = \displaystyle a+d+d      [from (i)]

                = \displaystyle a+2d

        or         = \displaystyle a+(3-1)d

    Fourth term \displaystyle {{a}_{4}}={{a}_{3}}+d

        or          = \displaystyle a+2d+d     [from (iii)]

                 = \displaystyle a+3d

                 = \displaystyle a+(4-1)d

    \      nth term an \displaystyle =a+(n-1)d

Note: The nth term of the A.P. with first term a & common difference d is given by \displaystyle {{a}_{n}}=a+(n-1)d

\displaystyle {{a}_{n}} is also called as general term of an A.P.

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

Problems based on nth TERM OF AN Arithmetic Progression

Problem:

Find the 18th term and nth term for the sequence 7, 4, 1, 2, 5.

Solution:

Here      a = 7

        and      d = \displaystyle {{a}_{2}}-{{a}_{1}}

             = 4 7 = 3

             n = 18

         \displaystyle {{a}_{n}}=a+\left( n-1 \right)d

         \displaystyle {{a}_{18}}=7+\left( 18-1 \right)\times -3

             = 7 + 17 ´ 3

             = 7 51

             = 44

            \displaystyle {{a}_{n}}=a+\left( n-1 \right)d

             = 7 + ( 1) (3)

             = 7 3n + 3

             = 10 3n

Problem:

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

Solution:

a = 7

             \displaystyle d={{a}_{2}}-{{a}_{1}}

              = 12 7 = 5

            \displaystyle {{a}_{n}}=87

        As     \displaystyle {{a}_{n}}=a+\left( n-1 \right)d

            87 = 7 + ( 1) ´ 5

         \displaystyle 87-7=5n-5

         \displaystyle 80+5=5n

            \displaystyle \frac{85}{5}=n

            17 = n

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

Problem:

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

Solution:

a = 7

         \displaystyle d={{a}_{2}}-{{a}_{1}}

         = 13 7 = 6

         \displaystyle {{a}_{n}}=a+\left( n-1 \right)d

        \displaystyle 205=7+\left( n-1 \right)\times 6

        205 7 = 6n 6

        198 + 6 = 6n

         \displaystyle \frac{204}{6}=n 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

        \displaystyle d={{a}_{2}}-{{a}_{1}}

         = 8 11 = -3

        let \displaystyle {{a}_{n}}=-150                [Assume]

        \displaystyle -150-11=-3n+3

        \displaystyle -161-3=-3n

        \displaystyle -164=-3n

         \displaystyle \frac{164}{3}=n

        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,  ,  , \displaystyle 9\frac{1}{2}            (ii) 4,  ,  ,  ,  , 6

Solution:

(i)    Here a = 5

            & \displaystyle {{a}_{4}}=9\frac{1}{2} or \displaystyle \frac{19}{2}

        Now,      \displaystyle {{a}_{4}}=a+3d

             \displaystyle \frac{19}{2}=5+3d

            \displaystyle \frac{19}{2}-5=3d

             \displaystyle \frac{9}{2}=3d

             \displaystyle \frac{9}{6}=d

        \     \displaystyle d=\frac{3}{2}

        \     \displaystyle {{a}_{2}}=a+d=5+\frac{3}{2}=\frac{13}{2}

             \displaystyle {{a}_{3}}={{a}_{2}}+d=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=8

        Hence     

        (ii)     Here \displaystyle a=-4

            \displaystyle {{a}_{6}}=6

        As    \displaystyle {{a}_{6}}=a+5d

             6 = 4 + 5d

            \displaystyle \frac{10}{5}=d

             d = 2

        \     \displaystyle {{a}_{2}}=a+d=-4+2=-2

            \displaystyle {{a}_{3}}={{a}_{2}}+d=-2+2=0

            \displaystyle {{a}_{4}}={{a}_{3}}+d=0+2=2

            \displaystyle {{a}_{5}}={{a}_{4}}+d=2+2=4

        

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

            \displaystyle {{a}_{n}}=63+\left( n-1 \right)2

             = 63 + 2n2

             = 61 + 2n

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

        \ \displaystyle b=3,\,\,\,{{d}_{2}}=7

        \ \displaystyle {{b}_{n}}=3+\left( n-1 \right)7

         =     \displaystyle 3+7n-7

         = 7n4

        According to question

        61 + 2n = 7n 4

        61 + 4 = 7n 2n

          65 = 5n

         \displaystyle \frac{65}{5}=n

        \n = 13

Problem:

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

Solution:

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

        Then, their 100th terms will be

            \displaystyle {{a}_{100}}={{a}_{1}}+99d;\,\,\,\,\,{{b}_{100}}={{b}_{1}}+99d

        According to question     \displaystyle {{a}_{100}}-{{b}_{100}}=100

        i.e.            …(i)

        Now, difference between their 1000th terms

            \displaystyle ={{a}_{1000}}-{{b}_{1000}}=\left( {{a}_{1}}+999d \right)-\left( {{b}_{1}}+999d \right)

            \displaystyle ={{a}_{1}}-{{b}_{1}}=100        [By equation (i)]

Problem:

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

Solution:

The give A.P. is 3, 10, 17, ……

        Here, a (first term)     = 3

        d (common difference)= 10 3 = 7

                 \displaystyle {{a}_{n}}=a+\left( n-1 \right)d

                 \displaystyle {{a}_{13}}=3+(13-1)7

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

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

        So, we get     \displaystyle {{a}_{n}}=84+{{a}_{13}}

             \displaystyle a+\left( n-1 \right)d=84+a+12d

            \displaystyle d\left( n-1-12 \right)=84

             \displaystyle 7\left( n-13 \right)=84

         \displaystyle n-13=\frac{84}{7}=12

         \displaystyle n=12+13=25

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

Problem:

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

Solution:

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

        Here    \displaystyle {{a}_{11}}=a+10d=38                …(i)

            \displaystyle {{a}_{16}}=a+15d=73                …(ii)

        Subtracting (ii) and (i), we get

            \displaystyle a+10d-a-15d=38-73

    or            

        Putting d = 7 in (i), we get

            \displaystyle a+10\times 7=38

            \displaystyle a=38-70=-32

        \     \displaystyle {{a}_{31}}=a+30d=-32+30\times 7=-32+210=178

        Hence, 31st 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,      \displaystyle a=105,\,\,\,\,\,d=7,\,\,\,\,\,{{a}_{n}}=994

        \     \displaystyle a+(n-1)d=994

            \displaystyle 105+(n-1)7=994

        Þ     \displaystyle 7(n-1)=994-105=889

        Þ         \displaystyle n-1=\frac{889}{7}

        Þ         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 = \displaystyle \frac{P\times R\times T}{100}

        So, the interest at the end of the 1st year     = Rs \displaystyle \frac{1000\times 8\times 1}{100} = Rs 80

        The interest at the end of the 2nd year     = Rs \displaystyle \frac{1000\times 8\times 2}{100} = Rs 160

        The interest at the end of the 3rd year     = Rs \displaystyle \frac{1000\times 8\times 3}{100} = Rs 240

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

        So, the interest (in Rs) at the end of the 1st, 2nd, 3rd, ….. 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 \displaystyle {{a}_{30}}.

        Now, \displaystyle {{a}_{30}}=a+(30-1)d=80+29\times 80=2400

        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. \displaystyle a,a+d,\,a+2d,....\left( l-2d \right),\left( l-d \right), \displaystyle l

    where l is the last term

    last term          \displaystyle l = \displaystyle l (1 1)d

    2nd last term     \displaystyle l
d = \displaystyle l (2 1) d

    3rd last term      \displaystyle l 2d = \displaystyle l (3 1)d

            ……………………………

            ……………………………

Note: nth term from the end = \displaystyle l  (n 1) d

Problem:

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

Solution:

1st method

        \displaystyle l=-40,d=14-17=-3

        Using \displaystyle l-(n-1)d

        5th term from the end will be

        \displaystyle =-40-(5-1)\times -3

        \displaystyle =-40-4\times -3

        \displaystyle =-40+12

        = 28

2nd method

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

        \     \displaystyle a=-40

            \displaystyle d=-37-(-40)

             =37 + 40

             = 3

         n = 5

        Using \displaystyle {{a}_{n}}=a+(n-1)d

             =\displaystyle -40+(5-1)\times 3

             \displaystyle =-40+4\times 3

             = 40 + 12

             = 28

Problem:

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

Solution:

Using 1st condition

        \     \displaystyle {{a}_{4}}+{{a}_{8}}=24

            \displaystyle a+3d+a+7d=24

            \displaystyle 2a+10d=24

        or     \displaystyle a+5d=12            …(i)

        Using 2nd condition

             \displaystyle {{a}_{6}}+{{a}_{10}}=44

        or    \displaystyle a+5d+a+9d=44

            \displaystyle 2a+14d=44

        or    \displaystyle a+7d=22            …(ii)

        Subtracting equation (i) from (ii) we have

            \displaystyle 2d=10

        or     d = 5

        Putting value of d in equations (i)

            \displaystyle a+5\times 5=12

            \displaystyle a=12-25

            \displaystyle a=-13

        \ \displaystyle {{a}_{2}}=a+d

             \displaystyle =-13+5=-8

         \displaystyle {{a}_{3}}=a+2d

             \displaystyle =-13+2\times 5

             \displaystyle =-13+10=-3

Problem:

If the pth term of an A.P. is q and the qth term is p, prove that its nth term is \displaystyle \left( p+q-n \right).

Solution:

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

        Then,     pth term = q    Þ    \displaystyle a+\left( p-1 \right)d=q    …(i)

            qth term = p    Þ    \displaystyle a+\left( q-1 \right)d=p    …(ii)

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

                    …(iii)

        Putting \displaystyle d=-1 in equation (i), we get

            \displaystyle a+1-p=q

        

        \    nth term \displaystyle =a+\left( n-1 \right)d

            \displaystyle =\left( p+q-1 \right)+\left( n-1 \right)\times (-1)     [from equation (iii)]

            \displaystyle =\left( p+q-n \right)

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    \displaystyle m{{a}_{m}}=n{{a}_{n}}

            Þ    \displaystyle m\left\{ a+\left( m-1 \right)d \right\}=n\left\{ a+\left( n-1 \right)d \right\}

            Þ    \displaystyle m\left\{ a+\left( m-1 \right)d \right\}-n\left\{ a+\left( n-1 \right)d \right\}=0

            Þ    \displaystyle a\left( m-n \right)+\left\{ m\left( m-1 \right)-n\left( n-1 \right) \right\}d=0

            Þ    \displaystyle a\left( m-n \right)+\left( {{m}^{2}}-{{n}^{2}}-m+n \right)d=0

            Þ    \displaystyle a\left( m-n \right)+\left[ \left( m-n \right)\left( m+n \right)-\left( m-n \right) \right]d=0

            Þ    \displaystyle a\left( m-n \right)+\left[ \left( m-n \right)\left( m+n-1 \right)d \right]=0

            Þ    \displaystyle \left( m-n \right)\left[ a+\left( m+n-1 \right)d \right]=0

             Þ    \displaystyle a+\left( m+n-1 \right)d=0        [\displaystyle \because m\ne n]

            Þ    \displaystyle {{a}_{m+n}}=0

        \     \displaystyle \left( m+n \right)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,

        \displaystyle b-a= common difference = \displaystyle c-b

    Þ    \displaystyle b-a=c-b

    Þ    \displaystyle 2b=a+c

Note: a, b, c are in A.P. iff \displaystyle 2b=a+c

Problem:

If \displaystyle 2x,x+10,3x+2 are in A.P., find the value of x.

Solution:

Since, \displaystyle 2x,x+10,3x+2 are in A.P.

        \     \displaystyle 2\left( x+10 \right)=2x+\left( 3x+2 \right)

        Þ     \displaystyle 2x+20=5x+2

        Þ     \displaystyle 3x=18

        Þ     x = 6

Problem:

If the numbers a, b, c, d, e form an A.P., then find the value of \displaystyle a-4b+6c-4d+e.

Solution:

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

            \displaystyle b=a+D,c=a+2D,d=a+3D and \displaystyle e=a+4D

        \    \displaystyle a-4b+6c-4d+e=a-4(a+D)+6(a+2D)-4(a+3D)+(a+4D)

        Þ    \displaystyle a-4b+6c-4d+e=a-4a-4D+6a+12D-4a-12D+a+4D

        Þ    \displaystyle a-4b+6c-4d+e=a-4a+6a-4a+a-4D+12D-12D+4D

        Þ    \displaystyle a-4b+6c-4d+e=0

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