**Chapter 2 Sequences and Series Miscellaneous Exercise 2**

## Chapter 2 Sequences and Series Miscellaneous Exercise 2

**(I) Select the correct answer from the given alternative:**

**Question 1.The common ratio for the G.P. 0.12, 0.24, 0.48, is**

(A) 0.12

(B) 0.2

(C) 0.02

(D) 2

**Answer:**

(D) 2

**Question 2.**

**Answer:**

(C) -128

Hint:

**Question 3.(A) 3(B) 2(C) 1(D) -1Answer:**

(A) 3

Hint:

**Question 4.Which term of the geometric progression 1, 2, 4, 8, ….. is 2048.(A) 10th(B) 11th(C) 12th(D) 13thAnswer:**

(C) 12th

Hint:

Here, a = 1, r = 2

**Question 5.If the common ratio of the G.P. is 5, the 5th term is 1875, the first term is(A) 3(B) 5(C) 15(D) -5Answer:**

(A) 3

**Question 6.The sum of 3 terms of a G.P. is 21/4 and their product is 1, then the common ratio is(A) 1(B) 2(C) 4(D) 8Answer:**

(C) 4

Hint:

**Question 7.**

**Question 9.Which of the following is not true, where A, G, H are the AM, GM, HM of a and b respectively, (a, b > 0)Answer:**

(D) A = GH

**Question 10.The G.M. of two numbers exceeds their H.M. by 6/5, the A.M. exceeds G.M. by 3/2 the two numbers areAnswer:**

(C) 3, 12

Hint:

(II) Answer the following:

**Question 1.In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.Solution:**

**Question 2.Find the sum of the first 5 terms of the G.P. whose first term is 1 and the common ratio is 2/3.Solution:**

**Question 3.Solution:**

**Question 4.**

**Question 5.Find three numbers in G.P. such that their sum is 35 and their product is 1000.Solution:**

Substituting the value of a in (i), we get

Hence, the three numbers in G.P. are 20, 10, 5, or 5, 10, 20.

**Question 6.Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.Solution:**

According to the given condition,

**Question 7.For a sequence, S _{n} = 4( – 1), verify that the sequence is a G.P.Solution:**

∴ The given sequence is a G.P.

**Question 8.Find 2 + 22 + 222 + 2222 + … upto n terms.Solution:**

**Question 9.Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,…Solution:**

**Question 10.Solution:**

**Question 11.Solution:**

**Question 12.Solution:**

We know that

**Question 13.Solution:**

**Question 14.Find 2 × 6 + 4 × 9 + 6 × 12 + ….. upto n terms.Solution:**

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)

**Question 15.Find 2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + …… upto n terms.Solution:**

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)( + 7n + 12)

= 2n (n + 1) (n + 3) (n + 4)

**Question 16.Solution:**

**Question 17.Solution:**

**Question 18.Solution:**

**Question 19.Solution:**

**Question 20.Solution:**

**Question 21.For a G.P. if = 7, = 1575, find a.Solution:**

**Question 22.Solution:**

**Question 23.Solution:**

**Question 24.Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.Solution:**
Since k – 1, k, k + 2 are consecutive terms of a G.P.,

k – 2 = 0

∴ k = 2

**Question 25.If for a G.P. first term is (27 and the seventh term is (8, find .Solution:**

**Question 26.If pth, qth and rth terms of a G.P. are x, y, z respectively. Find the value of .Solution:**

Let a be the first term and R be the common ratio of the G.P.

**Question 27.Which 2 terms are inserted between 5 and 40 so that the resulting sequence is G.P.Solution:**

Let the required numbers be G

_{1}and G

_{2}.

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

**Question 28.Solution:**

**Question 29.**

**Solution:**

a, b, c are in G.P.

∴ = ac

a + 2bx + c = 0 becomes

**Question 30.Solution:**

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

**Question 31.**

i.e., we have to show

**Question 32.Find the coefficient in the expression of using series expansion.Solution:**

**Question 33.Solution:**