Sequences and Series

1 )

If the 1^st term and common ratio of a G. P. are 1 and 2 respectively then s₁ + s₃ + s₅ +…+ s_2n–1 =_______

(1/3)(22n – 5n + 4)

(1/3)(22n+1 – 5n)

(1/3)(22n+1 – 3n – 2)

(1/3)(22n+1 – 5n2)

View Solution

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)

2 )

(1/2) + (3/4) + (7/8) + (15 / 16) ...100 terms =_________

2^(100) + 99

2^(–100) + 99

2^(–101) + 100

2^(-99) + 99

View Solution

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

3 )

If for the triangle whose perimeter is 37 cms and length of sides are in G. P. also the length of the smallest side is 9 cms then length of remaining two sides are____ and ____

12, 16

14, 14

10, 18

15, 13

View Solution

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.

4 )

Find a, b and c between 2 and 18 such that a + b + c = 25, 2, a, b are in A. P. and b, c, 18 are in G. P.

5, 8, 12

4, 8, 13

3, 9, 13

5, 9, 11

View Solution

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

5 )

Find out four numbers such that, first three numbers are in G. P., last three numbers are in A. P. having common difference 6, first and last numbers are same.

8, 4, 2, 8

– 8, 4, – 2, – 8

8, – 4, 2, 8

– 8, – 4, – 2, – 8

View Solution

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

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