Chapter 2 Sequences and Series Miscellaneous Exercise 2 Solutions

(D) 2

(C) -128 Hint:

(A) 3 Hint:

(C) 12th Hint: Here, a = 1, r = 2 nth term of geometric progression = arn-1 ∴ arn-1 = 2048 2n-1 = 211 n – 1 = 11 ∴ n = 12

(A) 3

(C) 4 Hint: Let three terms be , a, ar According to the given conditions, + a + ar = …..(i) and × a × ar = 1, i.e., a3 = 1 ∴ a = 1 ∴ from equation (i), we get + 1 + r = By solving this, we get r = 4.

(C) Hint: Here, a = 5, r = , |r| < 1 ∴ Sum to the infinity =

(A) Hint:

(D) A = GH

(C) 3, 12 Hint:

The sequence (tn) is a G.P. if = constant for all n ∈ N.

Let the three numbers in G.P. be , a, ar. According to the given conditions, + a + ar = 35 a( + 1 + r) = 35 …..(i) Also, ()(a)(ar) = 1000 a3 = 1000 ∴ a = 10 Substituting the value of a in (i), we get Hence, the three numbers in G.P. are 20, 10, 5, or 5, 10, 20.

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

∴ The given sequence is a G.P.

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 + ….. to n terms = n(n + 1) [2n + 1 + 3] = 2n(n + 1)(n + 2)

2, 4, 6,… are in A.P. ∴ rth term = 2 + (r – 1) 2 = 2r 5, 7, 9, … are in A.P. ∴ rth term = 5 + (r – 1) (2) = (2r + 3) 8, 10, 12, … are in A.P. ∴ rth term = 8 + (r – 1) (2) = (2r + 6) 2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + ….. to n terms = 2n (n + 1) [n(n + 1) + 3(2n + 1) + 9] = 2n (n + 1)(n2 + 7n + 12) = 2n (n + 1) (n + 3) (n + 4)

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.

Let the required numbers be G1 and G2. ∴ For the resulting sequence to be in G.P. we need to insert numbers 10 and 20.

a, b, c are in G.P. ∴ b2 = ac ax2 + 2bx + c = 0 becomes

p, q, r, s are in G.P.

p, q, r, s are in G.P. Let the common ratio be R ∴ let p = , q = , r = aR and s = aR3 To show that (pn + qn), (qn + rn), (rn + sn) are in G.P, i.e., we have to show