Sequences%20and%20Series

Answer :

given: a = 1, r = 2 for G. P.

          ∴      s₁ + s₃ + s₅ + --- s_2n–1  = 1 + 1[(1 – 2³) / (1 – 2)] + 1[(1 – 2⁵) / (1 – 2)] + ---- n terms

                   = 1 + 1(2³ – 1) + (2⁵ – 1) + ---- n terms

                   = (2 + 2³ + 2⁵ + --- n terms) – (1 + 1 + 1 --- n terms)

                   = 2 ∙ [(2²ⁿ – 1) / (2² – 1)] – n

                   = (2/3)(2²ⁿ – 1) – n

                   = [(2²ⁿ⁺¹ – 2) / 3] – n

                   = [(2²ⁿ⁺¹ – 2 – 3n) / 3]

                   = (1/3)(2²ⁿ⁺¹ – 3n – 2)

Answer :

  Required sum = (1/2) + (3/4) + (7/8) + (15 / 16) + --- 100 terms

               = [1 – (1/2)] + [1 – (1/4)] + [1 – (1/8)] + --- 100 terms

               = (1 + 1 + 1 + --- 100 terms) – [(1/2) + (1/4) + (1/8) + --- 100 terms]

               = 100 – (1/2)[{1 – (1/2)¹⁰⁰} / {1 – (1/2)}]

               = 100 – 1[1 – (1/2)¹⁰⁰]

               = 100 – 1 + (1/2)¹⁰⁰

               = 2^–100 + 99

Answer :

Length of sides of triangle in G. P. Let side be 9, 9r, 9r² (r > 1) Perimeters = 37

          ∴       9 + 9r + 9r² = 37

          ∴       9r² + 9r – 28 = 0

          ∴       (3r + 7) (3r – 4) = 0

          ∴       r = [(– 7) / 3] and r = (4/3)

                   but r > 0 hence r = (4/3)

          ∴       sides are : 9, 9r, 9r² i.e. 9, 12, 16

          ∴       other two sides are 12 and 16.

Answer :

from given condition, a + b + c = 25                    (1)

          2a = 2 + b                                                             (2)

          c² = 18b                                                               (3)

          by (1) & (2), we get, [(a + b) / 2] + b + c = 25 ⇒ 3b + 2c = 48          (4)

          by (3) & (4), we get, 3[c² / (18)] + 2c = 48

          ∴       c² + 12c – 288 = 0

          ∴       (c – 12) (c + 24) = 0

          ∴       c = 12 or c = – 24

          ∴       c = 12 as c ≠ negative

          as c = 12, from (3), b = [(144) / (18)] = 8

          from (2) a = [(2 + b) / 2] = 5

          ∴       a = 5, b = 8 and c = 12

Answer :

Let     numbers be a + 6, a – 6, a, a + 6

                   from given condition, (a – 6)² = (a + 6) (a)

          ∴       a² – 12a + 36 = a² + 6

          ∴       36 = 18

          ∴       a = 2

          ∴       Numbers are 8, – 4, 2, 8