17 + Interview Questions in GEOMETRIC PROGRESSION IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Find the required term of the following sequence using sequence formula:(i) 2, 10, 50, … (6th term)(ii) 100, 50, 25, … (7th term)
Answer» (i) 2, 10, 50, … (6th term) Here. a = 2, r = 10/2 = 5, n = 6 T₆ = a.r^n-1 ∴ T₆ = 2.(5)^6-1 = 2 (5)⁵ = 2 × 3125 = 6250 (ii) 100, 50, 25, … (7th term) Here, a = 100; r = 50/100 = 1/2 = n = 7 T_n = a.r^n-1 ∴T₇ = \(100(\frac{1}{2})^7−1\) = \(100(\frac{1}{2})^6\) = 100/64 = 25/16

1.

Find the required term of the following sequence using sequence formula:(i) 2, 10, 50, … (6th term)(ii) 100, 50, 25, … (7th term)

Answer»

(i) 2, 10, 50, … (6th term)

Here. a = 2, r = 10/2 = 5, n = 6

T₆ = a.r^n-1

∴ T₆ = 2.(5)^6-1

= 2 (5)⁵

= 2 × 3125

= 6250

(ii) 100, 50, 25, … (7th term)

Here, a = 100; r = 50/100 = 1/2 = n = 7

T_n = a.r^n-1

∴T₇ = \(100(\frac{1}{2})^7−1\)

= \(100(\frac{1}{2})^6\)

= 100/64 = 25/16

Discussion

2.	What is the nth term of the G.P. ar, ar2, ar3 …?
Answer» The nth term of the G.P. ar, ar², ar³, … is a.rⁿ.

Discussion

3.	Find the sum of twenty terms of the G.P. 7, 7, 7,….
Answer» G.P. 7, 7, 7, … Here, a = 7; r = 1; n = 20. S_n = na S₂₀ = 20 × 7 = 140

3.

Find the sum of twenty terms of the G.P. 7, 7, 7,….

Answer»

G.P. 7, 7, 7, …

Here, a = 7; r = 1; n = 20. S_n = na

S₂₀ = 20 × 7 = 140

Discussion

4.	For a G.P., T1 = 2 and the product of the first three terms is 1000. Find the common ratio.
Answer» For a G.P., T₁ = 2 ∴ a = 2 Now, T₁ × T₂ × T₃ = a × ar × ar² ∴ 1000 = 2 × 2r × 2r³ ∴ 1000 = 8r³ ∴ r³ = 1000/8 = 125 = (5)³ (∴ Index are equal then base are equal) Hence, common ratio r = 5

4.

For a G.P., T1 = 2 and the product of the first three terms is 1000. Find the common ratio.

Answer»

For a G.P., T₁ = 2

∴ a = 2

Now, T₁ × T₂ × T₃ = a × ar × ar²

∴ 1000 = 2 × 2r × 2r³

∴ 1000 = 8r³

∴ r³ = 1000/8 = 125 = (5)³

(∴ Index are equal then base are equal)

Hence, common ratio r = 5

Discussion

5.	Find the common ratio of the G.P. 0. 1, 0.01, 0.001
Answer» The common ratio of the G.P. 0.1, 0.01, 0.001 is r = \(\frac{0.01}{0.1}\) = 0.1.

Discussion

6.	Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Answer» The incomes that Subba Rao earned is in A.P., as every year his salary is increased by Rs 200. Therefore, the salaries of each year from 1995 5000, 5200, 5400, … Here, a = 5000 d = 200 Let after n^th year, his salary be Rs 7000. Therefore, a_n = a + (n − 1) d 7000 = 5000 + (n − 1) 200 200(n − 1) = 2000 (n − 1) = 10 n = 11 Therefore, in 11th year, his salary will be Rs 7000.

6.

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Answer»

The incomes that Subba Rao earned is in A.P., as every year his salary is increased by Rs 200.

Therefore, the salaries of each year from 1995

5000, 5200, 5400, …

Here, a = 5000

d = 200

Let after n^th year, his salary be Rs 7000.

Therefore, a_n = a + (n − 1) d

7000 = 5000 + (n − 1) 200

200(n − 1) = 2000

(n − 1) = 10

n = 11

Therefore, in 11th year, his salary will be Rs 7000.

Discussion

7.	If a is the A.M. of b and c and the two geometric means are G1 and G2, then \(G_1^3\) + \(G_2^3\) is equal to (a) \(\frac{abc}{2}\)(b) abc (c) 2abc (d) \(\frac{3}{2}abc\)
Answer» (c) 2abc a is the A.M of b and c ⇒ 2a = b + c Given, G₁ and G₂ are G.Ms between b and c ⇒ b, G₁, G₂, c are in G.P. Let r be the common ratio of the G.P. ⇒ G₁ = br, G₂ = br², c = br³ Now c = br³ ⇒ r³ = \(\frac{c}{b}\) ⇒ r = \(\big(\frac{c}{b}\big)^\frac{1}{3}\) ∴ G₁ = b x \(\frac{c^\frac{1}{3}}{b^\frac{1}{3}}\) = \(b^\frac{2}{3}\) \(c^\frac{1}{3}\) = \((b^2c)^\frac{1}{3}\) G₂ = b x \(\big(\frac{c}{b}\big)^\frac{2}{3}\) = \(\frac{b\times{c}^\frac{2}{3}}{b^\frac{2}{3}}\) = \(b^\frac{1}{3}\) \(c^\frac{2}{3}\) = \((bc^2)^\frac{1}{3}\) Now \(G_1^3\) + \(G_2^3\) = \(\big((b^2c)^\frac{1}{3}\big)^3\) + \(\big((bc^2)^\frac{1}{3}\big)^3\) = b²c + bc² = bc (b + c) = bc . 2a = 2abc.

7.

If a is the A.M. of b and c and the two geometric means are G1 and G2, then \(G_1^3\) + \(G_2^3\) is equal to (a) \(\frac{abc}{2}\)(b) abc (c) 2abc (d) \(\frac{3}{2}abc\)

Answer»

(c) 2abc

a is the A.M of b and c ⇒ 2a = b + c

Given, G₁ and G₂ are G.Ms between b and c

⇒ b, G₁, G₂, c are in G.P.

Let r be the common ratio of the G.P.

⇒ G₁ = br, G₂ = br², c = br³

Now c = br³ ⇒ r³ = \(\frac{c}{b}\) ⇒ r = \(\big(\frac{c}{b}\big)^\frac{1}{3}\)

∴ G₁ = b x \(\frac{c^\frac{1}{3}}{b^\frac{1}{3}}\) = \(b^\frac{2}{3}\) \(c^\frac{1}{3}\) = \((b^2c)^\frac{1}{3}\)

G₂ = b x \(\big(\frac{c}{b}\big)^\frac{2}{3}\) = \(\frac{b\times{c}^\frac{2}{3}}{b^\frac{2}{3}}\) = \(b^\frac{1}{3}\) \(c^\frac{2}{3}\) = \((bc^2)^\frac{1}{3}\)

Now \(G_1^3\) + \(G_2^3\) = \(\big((b^2c)^\frac{1}{3}\big)^3\) + \(\big((bc^2)^\frac{1}{3}\big)^3\)

= b²c + bc² = bc (b + c) = bc . 2a = 2abc.

Discussion

8.	If x, 10, -25 are in G.P, then find the value of x.(a) 4(b) – 25(c) – 4(d) 2
Answer» Correct option is (c) – 4

Discussion

9.	If x > 0 and log3x + log3 \(\sqrt{x}\) + \(\text{log}_3 (\sqrt[4]x)\) + \(\text{log}_3 (\sqrt[8]x)\)+ \(\text{log}_3 (\sqrt[16]x)\)+ ..... = 4, then x equals(a) 1 (b) 9 (c) 27 (d) 81
Answer» (b) 9. log₃x + log₃ \(\sqrt{x}\) + \(\text{log}_3 (\sqrt[4]x)\) + \(\text{log}_3 (\sqrt[8]x)\)+ ..... \(\infty\) = 4 ⇒ log₃x + log₃x^1/2 + log₃x^1/4+ log₃ (x^1/8) + ..... ∞ = 4 ⇒ log₃ (x . x^1/2 . x^1/4 . x^1/8 ..... ∞) = 4 ⇒ log₃ (x^{1 + 1/2 + 1/4 + 1/8} ..... ∞) = 4 ⇒ log₃\(x^{\frac{1}{1-\frac{1}{2}}}\) = 4 (∞ S_∞ = \(\frac{a}{r}\) – r) ⇒ log₃ x² = 4 ⇒ x² = 3⁴ = (3²)² ⇒ \(x\) = 9.

9.

If x > 0 and log3x + log3 \(\sqrt{x}\) + \(\text{log}_3 (\sqrt[4]x)\) + \(\text{log}_3 (\sqrt[8]x)\)+ \(\text{log}_3 (\sqrt[16]x)\)+ ..... = 4, then x equals(a) 1 (b) 9 (c) 27 (d) 81

Answer»

(b) 9.

log₃x + log₃ \(\sqrt{x}\) + \(\text{log}_3 (\sqrt[4]x)\) + \(\text{log}_3 (\sqrt[8]x)\)+ ..... \(\infty\) = 4

⇒ log₃x + log₃x^1/2 + log₃x^1/4+ log₃ (x^1/8) + ..... ∞ = 4

⇒ log₃ (x . x^1/2 . x^1/4 . x^1/8 ..... ∞) = 4

⇒ log₃ (x^{1 + 1/2 + 1/4 + 1/8} ..... ∞) = 4

⇒ log₃\(x^{\frac{1}{1-\frac{1}{2}}}\) = 4 (∞ S_∞ = \(\frac{a}{r}\) – r)

⇒ log₃ x² = 4 ⇒ x² = 3⁴ = (3²)² ⇒ \(x\) = 9.

Discussion

10.	The common ratio of a G.P. is -1 and its first term is -1, then find the sum of the first six terms of the G.P.(a) 0(b) – 1(c) 1(d) 6
Answer» Correct option is (a) 0

Discussion

11.	The sum of the first three terms of a G.P. is \(\frac{39}{10}\), and their product is 1. Find the common ratio and the terms.
Answer» Let the three numbers be \(\frac{a}{r},a, ar\). ∴ According to the question \(\Rightarrow\) \(\frac{a}{r}\) + a+ ar = \(\frac{39}{10}\)...... (1) \(\Rightarrow\) \(\frac{a}{r}\) \(\times a \times ar = 1\) ...... (2) From 2 we get ⇒ a³ = 1 ∴ a = 1. From 1 we get \(\frac{a + ar + ar^2}{r} = \frac{39}{10}\) ⇒ 10a + 10ar + 10ar² = 39r …(3) Substituting a = 1 in 3 we get ⇒ 10(1) + 10(1)r + 10(1)r² = 39r ⇒10r² – 29r + 10 = 0 ⇒ 10r² – 25r – 4r + 10 = 0…(4) ⇒ 5r(2r – 5) – 2(2r – 5) = 0 \(\therefore\) r = \(\frac{2}{5}\) or r = \(\frac{5}{2}\) \(\therefore\) Now the equation will be ⇒ \(\cfrac{1}{\frac{2}{5}}\), 1,1\(\times\) \(\frac{2}{5}\) or \(\frac{1}{\frac{5}{2}}\),1,1 \(\times\) \(\frac{5}{2}\) ⇒ \(\frac{5}{2}, 1, \frac{2}{5}\) or \(\frac{5}{2}, 1, \frac{2}{5}\) \(\therefore\) The three numbers are \(\frac{5}{2}, 1, \frac{2}{5}\)

11.

The sum of the first three terms of a G.P. is \(\frac{39}{10}\), and their product is 1. Find the common ratio and the terms.

Answer»

Let the three numbers be \(\frac{a}{r},a, ar\).

∴ According to the question

\(\Rightarrow\) \(\frac{a}{r}\) + a+ ar = \(\frac{39}{10}\)...... (1)

\(\Rightarrow\) \(\frac{a}{r}\) \(\times a \times ar = 1\) ...... (2)

From 2 we get

⇒ a³ = 1

∴ a = 1. From 1 we get

\(\frac{a + ar + ar^2}{r} = \frac{39}{10}\)

⇒ 10a + 10ar + 10ar² = 39r …(3)

Substituting a = 1 in 3 we get

⇒ 10(1) + 10(1)r + 10(1)r² = 39r

⇒10r² – 29r + 10 = 0

⇒ 10r² – 25r – 4r + 10 = 0…(4)

⇒ 5r(2r – 5) – 2(2r – 5) = 0

\(\therefore\) r = \(\frac{2}{5}\) or r = \(\frac{5}{2}\)

\(\therefore\) Now the equation will be

⇒ \(\cfrac{1}{\frac{2}{5}}\), 1,1\(\times\) \(\frac{2}{5}\) or \(\frac{1}{\frac{5}{2}}\),1,1 \(\times\) \(\frac{5}{2}\)

⇒ \(\frac{5}{2}, 1, \frac{2}{5}\) or \(\frac{5}{2}, 1, \frac{2}{5}\)

\(\therefore\) The three numbers are \(\frac{5}{2}, 1, \frac{2}{5}\)

Discussion

12.	For a G.P., if S7 = 15 and S6 = 11, then find the seventh term of the G.P.
Answer» S₆ = 15, S₅ = 11 Now, T_n+1 = S_n+1 – S_n. We have to find 6th term. Therefore, put n = 5. ∴ T₆ = S₆ – S₅ = 15 – 11 = 4

12.

For a G.P., if S7 = 15 and S6 = 11, then find the seventh term of the G.P.

Answer»

S₆ = 15, S₅ = 11

Now, T_n+1 = S_n+1 – S_n. We have to find 6th term. Therefore, put n = 5.

∴ T₆ = S₆ – S₅ = 15 – 11 = 4

Discussion

13.	Which term of the G.P. 3, 3√3, 9, ...... is 2187 ?(a) 13 (b) 14 (c) 15 (d) 16
Answer» (a) 13 Here first term a = 3, common ratio r = \(\frac{3\sqrt3}{3}\) = \(\sqrt3\) Let the nth term be 2187. Then, T_n = ar^{n – 1} = 2187 ⇒ 3 x (\(\sqrt3\))^{n - 1}= 2187 ⇒ (\(\sqrt3\))^{n - 1}= 729 ⇒ (\(\sqrt3\))^{n - 1}= 3⁶ = (\(\sqrt3\))¹² ⇒ n – 1 = 12 ⇒ n = 13.

13.

Which term of the G.P. 3, 3√3, 9, ...... is 2187 ?(a) 13 (b) 14 (c) 15 (d) 16

Answer»

(a) 13

Here first term a = 3, common ratio r = \(\frac{3\sqrt3}{3}\) = \(\sqrt3\)

Let the nth term be 2187. Then,

T_n = ar^{n – 1} = 2187

⇒ 3 x (\(\sqrt3\))^{n - 1}= 2187

⇒ (\(\sqrt3\))^{n - 1}= 729

⇒ (\(\sqrt3\))^{n - 1}= 3⁶ = (\(\sqrt3\))¹²

⇒ n – 1 = 12 ⇒ n = 13.

Discussion

14.	If the first term of a G.P. is ‘a and the common ratio is ‘b’. Find the (n + 1)th term,(a) abn(b) arn(c)abn-1(d) arn-1
Answer» Correct option is (a) abⁿ

Discussion

15.	For a G.P. 1, √3, 3, 3√3 find the 5th term?(a) 9(b) 9√3(c) 27(d) \(\frac{(\sqrt3)^5-1}{(\sqrt3-1)}\)
Answer» Correct option is (a) 9

Discussion

16.	For a G.P. 1 / 9,1/ 3 1 …, find the seventh term.(a) 6561(b) 243(c) 81(d) 1 / 81
Answer» Correct option is (c) 81

Discussion

17.	For a G.P., T1 = a and T5 = 1/a: where, a > 0 then obtain Its third term.(a) a2(b) 1(c) \(\frac{(\sqrt3)^5-1}{\sqrt3-1}\)(d) a
Answer» Correct option is (b) 1

Discussion

Explore topic-wise InterviewSolutions in .

Find the required term of the following sequence using sequence formula:(i) 2, 10, 50, … (6th term)(ii) 100, 50, 25, … (7th term)

What is the nth term of the G.P. ar, ar2, ar3 …?

Find the sum of twenty terms of the G.P. 7, 7, 7,….

For a G.P., T1 = 2 and the product of the first three terms is 1000. Find the common ratio.

Find the common ratio of the G.P. 0. 1, 0.01, 0.001

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

If a is the A.M. of b and c and the two geometric means are G1 and G2, then \(G_1^3\) + \(G_2^3\) is equal to (a) \(\frac{abc}{2}\)(b) abc (c) 2abc (d) \(\frac{3}{2}abc\)

If x, 10, -25 are in G.P, then find the value of x.(a) 4(b) – 25(c) – 4(d) 2

If x &gt; 0 and log3x + log3 \(\sqrt{x}\) + \(\text{log}_3 (\sqrt[4]x)\) + \(\text{log}_3 (\sqrt[8]x)\)+ \(\text{log}_3 (\sqrt[16]x)\)+ ..... = 4, then x equals(a) 1 (b) 9 (c) 27 (d) 81

The common ratio of a G.P. is -1 and its first term is -1, then find the sum of the first six terms of the G.P.(a) 0(b) – 1(c) 1(d) 6

The sum of the first three terms of a G.P. is \(\frac{39}{10}\), and their product is 1. Find the common ratio and the terms.

For a G.P., if S7 = 15 and S6 = 11, then find the seventh term of the G.P.

Which term of the G.P. 3, 3√3, 9, ...... is 2187 ?(a) 13 (b) 14 (c) 15 (d) 16

If the first term of a G.P. is ‘a and the common ratio is ‘b’. Find the (n + 1)th term,(a) abn(b) arn(c)abn-1(d) arn-1

For a G.P. 1, √3, 3, 3√3 find the 5th term?(a) 9(b) 9√3(c) 27(d) \(\frac{(\sqrt3)^5-1}{(\sqrt3-1)}\)

For a G.P. 1 / 9,1/ 3 1 …, find the seventh term.(a) 6561(b) 243(c) 81(d) 1 / 81

For a G.P., T1 = a and T5 = 1/a: where, a &gt; 0 then obtain Its third term.(a) a2(b) 1(c) \(\frac{(\sqrt3)^5-1}{\sqrt3-1}\)(d) a

If x > 0 and log3x + log3 \(\sqrt{x}\) + \(\text{log}_3 (\sqrt[4]x)\) + \(\text{log}_3 (\sqrt[8]x)\)+ \(\text{log}_3 (\sqrt[16]x)\)+ ..... = 4, then x equals(a) 1 (b) 9 (c) 27 (d) 81

For a G.P., T1 = a and T5 = 1/a: where, a > 0 then obtain Its third term.(a) a2(b) 1(c) \(\frac{(\sqrt3)^5-1}{\sqrt3-1}\)(d) a