Chapter 4 Sequences and Series Miscellaneous Exercise 4 Solutions

The sequence (tn) is a G.P., if = constant, for all n ∈ N. ∴ the sequence is a G.P. with common ratio = ∴ first term = t1 =

Let the three numbers in G.P. be , a, ar. According to the first condition, ∴ the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.

Let the four numbers in G.P. be . According to the second condition, ∴ a4 = 1 ∴ a = 1 According to the first condition,

Let the five numbers in G.P. be According to the first condition,

Sn = 2 + 22 + 222 +….. upto n terms = 2(1 + 11 + 111 +…… upto n terms) = (9 + 99 + 999 + … upto n terms) = [(10 – 1) + (100 – 1) + (1000 – 1) +…… upto n terms] = [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)] Since, 10, 100, 1000, …… n terms are in G.P. with a = 10, r = = 10

0.6, 0.66, 0.666, 0.6666, …… ∴ t1 = 0.6 t2 = 0.66 = 0.6 + 0.06 t3 = 0.666 = 0.6 + 0.06 + 0.006 Hence, in general tn = 0.6 + 0.06 + 0.006 + …… upto n terms. The terms are in G.P.with a = 0.6, r = = 0.1 ∴ tn = the sum of first n terms of the G.P.

We know that,

2, 4, 6, … are in A.P. ∴ rth term = 2 + (r – 1)2 = 2r 6, 9, 12, … are in A.P. ∴ rth term = 6 + (r – 1) (3) = (3r + 3) ∴ 2 × 6 + 4 × 9 + 6 × 12 +…… upto n terms = n(n + 1) (2n + 1 + 3) = 2n(n + 1)(n + 2)

122 + 132 + 142 + 152 + …… + 202 = (12 + 22 + 32 + 42 + ……. + 202) – (12 + 22 + 32 + 42 + …… + 112) = 2870 – 506 = 2364

(502 – 492) + (482 – 472) + (462 – 452) + …… + (22 – 12) = (502 + 482 + 462 + …… + 22) – (492 + 472 + 452 + …… + 12) = = 1300 – 25 = 1275

Since k – 1, k, k + 2 are consecutive terms of a G.P. ∴ ∴ k2 = k2 + k – 2 ∴ k – 2 = 0 ∴ k = 2

Let a be the first term and R be the common ratio of the G.P. ∴ tn = ∴ x = , y = , z =