Table of Contents
Exercise 6.4 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on geometric progressions (GPs), identifying GPs, finding terms, and solving related problems. Mathematical expressions are rendered using MathJax.
1. In which of the following situations, does the list of numbers involved in the form of a GP?
(i) Salary of Sharmila, when her salary is ₹5,00,000 for the first year and expected to receive yearly increment of 10%.
First year: ₹5,00,000.
Second year: \( 5,00,000 \cdot (1 + 0.10) = 5,50,000 \).
Third year: \( 5,50,000 \cdot 1.10 = 5,00,000 \cdot (1.10)^2 = 6,05,000 \).
Sequence: 5,00,000, 5,50,000, 6,05,000, …
Ratio: \( \frac{5,50,000}{5,00,000} = 1.10 \), \( \frac{6,05,000}{5,50,000} = 1.10 \), constant ratio, so it forms a GP.
Forms a GP: Yes, Common ratio: 1.10
(ii) Number of bricks needed to make each step, if the stair case has total 30 steps, provided that bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.
Bottom step: 100 bricks.
Second step: \( 100 – 2 = 98 \).
Third step: \( 98 – 2 = 96 \).
Sequence: 100, 98, 96, …
Difference: \( 98 – 100 = -2 \), \( 96 – 98 = -2 \), constant difference, so it forms an AP, not a GP.
Forms a GP: No, Reason: Forms an AP with common difference -2
(iii) Perimeter of the each triangle, when the mid points of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid points in turn are joined to form still another triangle and the process continues indefinitely.
First triangle perimeter (equilateral, side 24 cm): \( 3 \cdot 24 = 72 \) cm.
Second triangle: Midpoints divide each side into 2 equal parts, so side = \( \frac{24}{2} = 12 \) cm, perimeter = \( 3 \cdot 12 = 36 \) cm.
Third triangle: Side = \( \frac{12}{2} = 6 \) cm, perimeter = \( 3 \cdot 6 = 18 \) cm.
Sequence: 72, 36, 18, …
Ratio: \( \frac{36}{72} = \frac{1}{2} \), \( \frac{18}{36} = \frac{1}{2} \), constant ratio, so it forms a GP.
Forms a GP: Yes, Common ratio: \frac{1}{2}
2. Write three terms of the GP when the first term \( a \) and the common ratio \( r \) are given?
(i) \( a = 4, r = 3 \)
First term: 4.
Second term: \( 4 \cdot 3 = 12 \).
Third term: \( 12 \cdot 3 = 36 \).
Terms: 4, 12, 36
(ii) \( a = \sqrt{5}, r = \frac{1}{5} \)
First term: \( \sqrt{5} \).
Second term: \( \sqrt{5} \cdot \frac{1}{5} = \frac{\sqrt{5}}{5} \).
Third term: \( \frac{\sqrt{5}}{5} \cdot \frac{1}{5} = \frac{\sqrt{5}}{25} \).
Terms: \sqrt{5}, \frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{25}
(iii) \( a = 81, r = -\frac{1}{3} \)
First term: 81.
Second term: \( 81 \cdot \left(-\frac{1}{3}\right) = -27 \).
Third term: \( -27 \cdot \left(-\frac{1}{3}\right) = 9 \).
Terms: 81, -27, 9
(iv) \( a = \frac{1}{64}, r = 2 \)
First term: \( \frac{1}{64} \).
Second term: \( \frac{1}{64} \cdot 2 = \frac{1}{32} \).
Third term: \( \frac{1}{32} \cdot 2 = \frac{1}{16} \).
Terms: \frac{1}{64}, \frac{1}{32}, \frac{1}{16}
3. Which of the following are GP? If they are in GP, write next three terms?
(i) 4, 8, 16, …
Ratio: \( \frac{8}{4} = 2 \), \( \frac{16}{8} = 2 \), constant, so it is a GP.
Common ratio \( r = 2 \).
Next terms: \( 16 \cdot 2 = 32 \), \( 32 \cdot 2 = 64 \), \( 64 \cdot 2 = 128 \).
Is a GP: Yes, Next terms: 32, 64, 128
(ii) \( \frac{1}{3}, -\frac{1}{6}, \frac{1}{12}, \ldots \)
Ratio: \( \frac{-\frac{1}{6}}{\frac{1}{3}} = -\frac{1}{2} \), \( \frac{\frac{1}{12}}{-\frac{1}{6}} = -\frac{1}{2} \), constant, so it is a GP.
Common ratio \( r = -\frac{1}{2} \).
Next terms: \( \frac{1}{12} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{24} \), \( -\frac{1}{24} \cdot \left(-\frac{1}{2}\right) = \frac{1}{48} \), \( \frac{1}{48} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{96} \).
Is a GP: Yes, Next terms: -\frac{1}{24}, \frac{1}{48}, -\frac{1}{96}
(iii) 5, 55, 555, …
Ratio: \( \frac{55}{5} = 11 \), \( \frac{555}{55} = 10.09 \), not constant, so not a GP.
Is a GP: No
(iv) \(-2, -6, -18, \ldots\)
Ratio: \( \frac{-6}{-2} = 3 \), \( \frac{-18}{-6} = 3 \), constant, so it is a GP.
Common ratio \( r = 3 \).
Next terms: \( -18 \cdot 3 = -54 \), \( -54 \cdot 3 = -162 \), \( -162 \cdot 3 = -486 \).
Is a GP: Yes, Next terms: -54, -162, -486
(v) \( \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \ldots \)
Ratio: \( \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \), \( \frac{\frac{1}{6}}{\frac{1}{4}} = \frac{2}{3} \), not constant, so not a GP.
Is a GP: No
(vi) 3, \(-3^2, 3^3, \ldots\)
Terms: 3, \(-9\), 27, …
Ratio: \( \frac{-9}{3} = -3 \), \( \frac{27}{-9} = -3 \), constant, so it is a GP.
Common ratio \( r = -3 \).
Next terms: \( 27 \cdot (-3) = -81 \), \( -81 \cdot (-3) = 243 \), \( 243 \cdot (-3) = -729 \).
Is a GP: Yes, Next terms: -81, 243, -729
(vii) \( x, 1, \frac{1}{x}, \ldots (x \neq 0) \)
Ratio: \( \frac{1}{x} \), \( \frac{\frac{1}{x}}{1} = \frac{1}{x} \), constant, so it is a GP.
Common ratio \( r = \frac{1}{x} \).
Next terms: \( \frac{1}{x} \cdot \frac{1}{x} = \frac{1}{x^2} \), \( \frac{1}{x^2} \cdot \frac{1}{x} = \frac{1}{x^3} \), \( \frac{1}{x^3} \cdot \frac{1}{x} = \frac{1}{x^4} \).
Is a GP: Yes, Next terms: \frac{1}{x^2}, \frac{1}{x^3}, \frac{1}{x^4}
(viii) \( \sqrt{2}, -2, 2\sqrt{2}, \ldots \)
Ratio: \( \frac{-2}{\sqrt{2}} = -\sqrt{2} \), \( \frac{2\sqrt{2}}{-2} = -\sqrt{2} \), constant, so it is a GP.
Common ratio \( r = -\sqrt{2} \).
Next terms: \( 2\sqrt{2} \cdot (-\sqrt{2}) = -4 \), \( -4 \cdot (-\sqrt{2}) = 4\sqrt{2} \), \( 4\sqrt{2} \cdot (-\sqrt{2}) = -8 \).
Is a GP: Yes, Next terms: -4, 4\sqrt{2}, -8
(ix) 0.4, 0.04, 0.004, …
Ratio: \( \frac{0.04}{0.4} = 0.1 \), \( \frac{0.004}{0.04} = 0.1 \), constant, so it is a GP.
Common ratio \( r = 0.1 \).
Next terms: \( 0.004 \cdot 0.1 = 0.0004 \), \( 0.0004 \cdot 0.1 = 0.00004 \), \( 0.00004 \cdot 0.1 = 0.000004 \).
Is a GP: Yes, Next terms: 0.0004, 0.00004, 0.000004
4. Find \( x \) so that \( x, x + 2, x + 6 \) are consecutive terms of a geometric progression.
For a GP, the ratio between consecutive terms is constant: \( \frac{x + 2}{x} = \frac{x + 6}{x + 2} \).
Cross-multiply: \( (x + 2)^2 = x (x + 6) \).
Expand: \( x^2 + 4x + 4 = x^2 + 6x \).
Simplify: \( x^2 + 4x + 4 – x^2 – 6x = 0 \implies -2x + 4 = 0 \implies 2x = 4 \implies x = 2 \).
Check: If \( x = 2 \), terms are 2, 4, 8. Ratio: \( \frac{4}{2} = 2 \), \( \frac{8}{4} = 2 \), forms a GP.
\( x = 2 \)
