1. Find d and write the next 4 terms of 21, 25, 29,-----

Ans :
     Common difference, d = t₂ - t₁
                          = 25 - 21 = 4.
Next 4 terms are   29 + 4 = 33,
                   33 + 4 = 37,
                   37 + 4 = 41,
                   41 + 4 = 45 .
2. Find d and write the next 4 terms of x + y, x - y, x - 3y, -----

Ans :
        Common difference, d = t₂ - t₁
                             = x-y - (x+y)
                             = x -y - x -y = -2y
  Next 4 terms are          x- 3y + (-2y) = x - 3y - 2y = x - 5y,
                            x - 5y + (-2y) = x - 5y - 2y = x - 7y,
                            x - 7y + (-2y) = x - 7y - 2y = x - 9y,
                            x - 9y + (-2y) = x - 9y - 2y = x - 11y.
3. Find t₁₀ and t_n for the A.P having a = 3 and d = 2 ?

Ans :
        t_n = a + (n - 1)d
           = 3 + (n - 1)2
           = 3 + 2n - 2 = 2n + 1
       t₁₀ = a + (10 - 1)d
           = 3 + 9 x 2 = 3+ 18
                     = 21 .

4. For an A. P a =   , d = 2. Find t₇ and t_n .

Ans :

5. Find the first five terms in the series whose n^th term is t_n = n (n+1) :

Ans :

      t₁ = 1( 1+1) = 2

     t₂  = 2 (2+1) = 2 x 3 = 6

     t₃  = 3 (3+1) = 3 x 4 = 12

     t₄  = 4 (4+1) = 4 x 5 = 20

     t₅  = 5 (5+1) = 5 x 6 = 30

Hence the required answer is 2, 6, 12, 20, 30 .

6. Find the first five terms in the series whose n^th term is t_n =

Ans :

7. Prove that the first two terms are equal to ' 0' and the rest are positive integers in a series if t_n = ( n - 1) ( n - 2) :

Ans :

        Given that t_n = ( n - 1 ) ( n -2 )

        Put n = 1, t₁ = (1 - 1 ) (1 - 2 )

                            = 0 x (-1) = 0.

            ie, first term = 0.

       Put n =2,  t₂ = (2 -1 ) ( 2 -2 )

                            = 1 x 0 = 0 .

             ie , second term = 0.

Hence the first two terms are ' 0 ' .

Since the n^th term is ( n - 1 ) ( n -2 ), the value of n greater than or equal to 3 are positive integers in a series .

8. Find the 17^th term in a series if  

Ans :

     Put n = 17, t₁₇

9. If t₁ = 1 and t_n = t_{n - 1} +3 for n ≥ 2 . Find the first 5 terms :

Ans :

      t₁ = 1

     t₂  = t₁ + 3

          = 1 + 3 = 4

     t₃  = t₂ + 3

          = 4 + 3 = 7

    t₄  = t₃ + 3 = 7 + 3 = 10.

   t₅  = t₄ + 3 = 10 + 3 = 13 .

Hence the first 5 terms are 1, 4, 7, 10, 13 .

10. Find the first 6 terms in the series t₁ = 1,  :

Ans :

11. Determine the 25^th term of an A.P whose 9^th term is -6 and common difference is :

Ans :

       Given that t₉ = ^-6

                 a + 8d = ^-6

      a + 10  = ^-6   a = ^-16

t₂₅  = a + 24d

            = -16 + 30 = 14 .

12. Each term of an  A . P is doubled . Is the resulting sequence also an A . P ? If it is , write its first term, common difference and n^th term :

Ans :

       General form of an A . P is a , a+d, a+2d, a+3d,------- If we double each term, we get

                 2a, 2 (a+d), 2 (a+2d), 2 (a+3d),---------

t₁ = 2a

t₂ = 2 (a+d)  = 2a +2d

t₃ = 2 (a+2d) = 2a + 4d

t₄ = 2 (a+3d) = 2a +6d

t₂ - t₁ = 2a + 2d - 2a = 2d

t₃ - t₂ = 2a + 4d - ( 2a +2d ) = 2d

t₄ - t₃ = 2a + 6d - (2a+4d ) = 2d

Since the consecutive terms differ by 2d, the resulting sequence is an A. P .

         ie,  t₁ = 2a and

          C.D   = 2d .

n^th term t_n  = 2 (a+ (n-1) d )

                   = 2a + 2 (n-1) d .

13. If 7 times the 7^th term of an A.P is equal to 11 times the 11^th term, show that the 18^th term of it is zero :

Ans :

       Given that 7.t₇ = 11t₁₁

            7 ( a+6d)  = 11 ( a+10d )

            7a + 42d = 11a + 110d

            11a - 7a + 110d - 42d = 0

            4a + 68d = 0 a + 17d = 0

            a + ( 18-1)d = 0

            t₁₈ = 0 .

14.If m times the m^th term of an A .P is equal to n times the n^th term, prove that the ( m+n )^th term of the A. P is zero :

Ans :

      Given that mt_m  = nt_n .

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

       ma + m (m-1) d  = na + n (n-1) d

       ma - na + [ m (m-1) - n ( n-1) ]d = 0.

       (m-n) a + ( m² - m - n² +n ) d = 0

       ( m-n) a + [ m² - n² - (m-n) ] d = 0

       (m-n ) a + [ (m+n) (m-n) - (m-n) ] d = 0

       (m-n) [ a + (m+n-1) d ] = 0

       a + ( m+n-1)d  = 0

       t _m+n = 0 .

15. Determine the 2^nd term and r^th term of an A.P whose 6^th term is 12 and 8^th term is 22 .

Ans :

       Given that t₆ = 12 and t₈ = 22

         ie, a + 5d = 12 and____________(1)

              a + 7d = 22       ___________(2)

(2) -(1)   2d  = 10

              d    = 5

from (1), a = 12 - 5d

                           = 12 - 5x5

                           = 12 - 25 = ^-13

         t₂ = a+d

             = ^-13 + 5

             = ^-8

       t_r    = a + (r-1)  d

             = ^-13 + (r-1)5

             = -13 + 5r - 5 = 5r - 18 .

16. Determine K so that K +2, 4K - 6 and 3K - 2 are the three consecutive terms of an A.P :

Ans :

       If a, b, c are three consecutive terms of an A.P,

                    8K - 12 = 4K 4K = 12

                                                    K = 3 .

17. Which term of the A.P 10, 8, 6, ------ is ^-28 ?

Ans :

        Given that a = 10, d = 8 -10 = ^-2 and

                                   t_n  = ^-28

a + ( n-1) d = -28

10 + ( n-1) (^-2) = ^-28

10 - 2n + 2 = -28 12 + 28 = 2n

                                   2n = 40

                                   n = 20 .

18. Find the sum of 16, 11, 6, -------- to 23 terms ; n terms :

Ans :

        Here a = 16, d = 11 - 16 = ^-5 , n = 23 .

          .

19. Find the sum of the arithmetic series x+y, x-y, x-3y, ------- to 22 terms ; P terms :

Ans :

       Here a = x + y , d = x - y - ( x+y) = ^-2y, n = 22 .

           S_n = [ 2a + ( n-1 )d ]

         S₂₂ = [ 2 x ( x + y) + (22 - 1 ) ( -2y) ]

                = 11 [ 2x + 2y - 42y ]

               = 11 ( 2x - 40y ) = 22 ( x - 20y )

        Sp  =   [ 2a + ( p-1) d ]

              = [ 2 ( x + y ) + ( p-1) ^-2y ]

              = ( 2x + 2y - 2py + 2y )

              = p ( x + y - py + y )

             = P ( x - py + 2y ) = P [ x - ( p-2) y ]

20. Find the sum of all natural numbers between 100 and 1000 which are multiples of 5 :

Ans :

       Here t₁ = 105 and

               t_n = 995

                d = 5

t_n = 995 a + (n-1)d = 995

              105 + (n-1) 5 = 995

              5n = 995 - 100 = 895

              .

21. Insert 5 arithmetic means between 4 and 22 :

Ans :

        Let 5 arithmetic means be a₁, a₂, a₃, a₄, a₅ .

The total number of terms in the series is 7

                      t₇  = 22

               a + 6d  = 22

               4 + 6d  = 22 6d = 18 d  = 3

The 5 arithmetic means are

             4 + 3, 4 + 6, 4 + 9,  4 +12, 4 + 15

             or 7, 10, 13, 16, 19 .

22. Find the sum to n terms in the series 1.2 + 2.3 + 3.4 + ---------

Ans :

       n^th term of this series is n (n+1) .

                 = ( n+1) ( n+2 ) .

23. If the sum of the first n natural numbers is S₁ and that of their squares S₂ and cubes S₃ . Show that  9 S²₂ = S₃ ( 1 + 8 S₁ ) _________ .

Ans :

From (1) and (2)

             9S²₂  = S₃ ( 1 + 8S₁ ) .

24. Find the common ratio and write the next four terms of the G . P   -3, 1, , --------- .

Ans :

        Given series is -3, 1, , --------

                      .

25. Find the common ratio and write the next four terms of the G . P , -----------

Ans :

26. Find t₄ and t_n of a G.P. with a = 1, r = 1.2 :

Ans :

       t₄  = ar³

            = 1 x ( 1.2 )³ = 1.728

       t_n  = ar ^n-1

            = 1 ( 1.2 ) ^n-1 = ( 1.2 ) ^n-1 .

27. Determine the 12^th term of a G . P whose 8^th term is 192 and common ratio is 2 .

Ans :

            Given that r = 2 and t₈ = 192

                          t₈ = ar⁷ = 192

                         a (2)⁷ = 192

                         a (2)⁷ = 2⁶ x 3

                        .

t₁₂ = ar¹¹

           = x (2)¹¹ = 3 x (2)¹⁰

                               = 3072 .

28. The 5^th, 8^th and 11^th terms of a G.P are p, q and s respectively . Show that q² = ps .

Ans :

       Given that ar⁴ = p  → (1)

                       ar⁷  = q and → (2)

                       ar¹⁰ = s       → (3)

29. Find the value of x so that are three consecutive terms of a G.P :

Ans :

30. Find the sum of the geometric progression, ,------- 10 terms :

Ans :

31. Evaluate ∑ ( 2 + 3^k) where k = 1, 2, ------ 11 .

Ans :

       ∑ ( 2 + 3^k ) where k = 1, 2, ------ 11

            = 2 + 2 + ---- 11 times + 3¹ + 3² + ---- + 3¹¹

32. The first term of a G.P is 2 and the sum to infinity is 6 . Find the common ratio .

Ans :

         Given that a = 2 and S_∞ = 6 .

We have

33. The common ratio of a G. P is , Find the first term :

Ans :