Telangana SCERT Solution Class X (10) Maths Chapter 6 Progressions Exercise 6.4
Exercise – 6.4
(Q1) Which of the following situations, does the list of numbers involved is in the form of a G.P?
(i) Salary of Sharmila: when her salary is RS 5,00,000 for the first year and expected to receive yearly increment of 10%
=> Solution:
Salary of Sharmila for the first year = RS 5,00,000
Her salary for the second year =
RS 500000 + 10%
= 500000 + 50000
= RS 550000
For 3rd year = RS 605000 and soon.
The list of her salaries, for 1st, 2nd and 3rd year.
RS 500000, RS 550000, RS 605000
Common ratio r = a2/a1 = 550000/500000
= r = a2/a1 = 11/10 = 1.1
r = a3/a2 = 605000/550000
= 11/10
r = 1.1
∴ It forms an G.P
(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 need 2 bricks less than the previous step.
=> Solution:
Total bricks needed for the first step = 100
For the second step = 100 – 2 = 98
For the third step = 98 – 2 = 96
∴ List of numbers involved in the situation, 100, 98, 96 ——
Common ratio r = a2/a1 = 98/100 = 49/50
∴ The list of the perimeters = 72, 36, 18, —–
Common ratio r = a2/a1 = 36/72 = 1/2
r = a3/a2 = 18/36 = 1/2
∴ a2/a1 = a3/a2
It forms an G.P
(Q2) Write three terms of the G.P when the first ‘a’ and the common ratio ‘r’ are given?
(i) a = 4, r = 3
=> Solution: The first term a1 = a = 4
a2 = ar = 4 X 3 = 12
a3 = a2r = 12 X 3 = 36
∴ The three terms are, 4, 12 and 36
(ii) a = √5, r = 1/5
=> Solution- = a = √5
a2 = ar = √5 X 1/5 = √5/5 = 1/√5
a3 = ar2 = √5 X (1/5)2 = √5/25
= √5/5X5
= √5/√5 X √5 X 5
= 1/5√5
∴ The three terms are,
√5, √5/5 and √5/25
(iii) a = 81, r = -1/3
= Solution: The first term a1 = a = 81
a2 = ar = 81 X (-1/3)
= -81/3 = -27
a3 = ar2 = 81 X (-1/3)2
= 81 X 1/9
= 9
∴ The three terms are, 81, -27 and 9
(iv) a = 1/64, r = 2
=> Solution: The first term a1 = a = 1/64
a2 = ar = 1/64 X2 = 1/32
a3 = ar2 = 1/64 X (2)2 = 1/64 X 4
= 1/16
∴ The three terms are 1/64, 1/32 and 1/16.
= -1/6 X 3/1
r = -1/2
r = a3/a2 = 1/12 X (-6/1) = -1/2
∴ a2/a1 = a3/a1
It forms a G.P
∴ The next three terms,
a4 = a3r = 1/12 X (-1/2) = -1/24
a5 = a4r= -1/24 X -1/2 = 1/48
a6 = a5r = 1/48 X (-1/2) = -1/96
∴ The next three terms = -1/24, 1/48 and -1/96
= -1/6 X 3/1
= -1/2
r = a3/a2 = 1/12 X (-6/1) = -1/2
∴ a2/a1 = a3/a1
It forms a G.P
∴ The next three terms,
a4 = a3r = 1/12 X (-1/2) = -1/24
a5 = a4r = -1/24 X -1/2 = 1/48
a6 = a5r = 1/48 X (-1/2) = -1/96
∴ The next three terms = -1/24, 1/48 and -1/96
(iii) 1/2, 1/4, 1/6
=> Solution: Each term is non-zero
Common ratio r = a2/a1 = (1/4)/(1/2) = 1/4 X 2/1
= 1/2
r = a3/a2 = (1/6)/(1/4) = 1/6 X 4/1 = 2/3
∴ a2/a1 ≠ a3/a2
∴ It doesn’t forms a G.P.
(iv) 5, 55, 555, —-
=> Solution: each term is non-zero
Common ratio, r = a2/a1 = 55/5 = 11
r = a3/a2 = 555/55 = 10.09 or 111/11
∴ a2/a1 ≠ a3/a2
∴ They are not in G.P
(v) -2, -6, -18,
=> Solution: Each term is non-zero
Common ratio, r = a2/a1 = -6/-2 = 3
r = a3/a2 = -18/-6 = 3
∴ It forms a G.P
The next three terms,
a4 = a3r = -18 X (3) = -54
a5 = a5r= -54 X 3 = -162
a6 = a5r = -162 X 3 = -486
∴ The next three terms = -54, -162 and -486
(vi) 3, -32, 33
=> Solution: each term is non-zero,
Common ratio, r = a2/a1 = -32/3
= -3
r = a3/a2 = 33/32 = -3
∴ a2/a1 = a3/a2
∴ It forms a G.P
The next three terms
a4 = a3r = 33 X (-3) = – 34
a5 = a4r = – 34 X (-3) = – 35
a6 = a5r = -35 X (-3) = – 36
∴ The next three terms = -34, -35 and -36
(vii) x, 1, 1/x, —— (x ≠ 0 )
=> Solution:
Each term is non-zero
Common ratio, r = a2/a1 = 1/x
r = a3/a2 = 11x/1 = 1/x
∴ a2/a1 = a3/a2
∴ They are in G.P
The next three terms,
a4 = a3r = 1/x X 1/x = 1/x2
a5 = a4r = 1/x2 X 1/x = 1/x3
a6 = a5r = 1/x3 X 1/x = 1/x4
∴ The next three terms = 1/x2, 1/x3 and 1/x4
(viii) 1/√2, -2, 4√2
=> Solution:
Each term is non-zero
Common ratio, r = a2/a1 = -2/1√2
= -2 X √2/1
= -2√2
r = a3/a2 = 4√2/-2 = -2√2
∴ a2/a1 = a3/a2
∴ They are in G.P
The next three terms,
a4 = a3r = 4√2 X -2√2 = -8 X 2 = -16
a5 = a4r = -16 X -2√2 = 32√2
a6 = a5r = 32√2 X -2√5 = -64 X 2 = 128
The next three terms = -16, 32√2 and – 128
(ix) 0.4, 0.04, 0.004, —-
=> Solution: each term is non-zero
Common ratio, r = a2/a1 = 0.04/0.4 = 0.1
r = a3/a2 = 0.004/0.04 = 0.1
∴ a2/a1 = a3/a2
∴ They are in G.P
The next three terms,
a4 = a3r = 0.004 X 0.1 = 0.0004
a5 = a4r = 0.0004 X 0.1 = 0.00004
a6 = a5r = 0.00004 X 0.1 = 0.000004
The next three terms = 0.0004, 0.00004, and 0.000004
(Q4) Find x so that, x, x+2, x+6 are consecutive terms of a G.P
=> Solution:
Let a1 = x, a2 = x + 2, and a3 = x + 6
Since they are in G.P
r = a2/a1 = a3/a2
=> x+2/x = x+6/x+2
=> (x + 2)2 = x (x + 6)
=> x2 + 4x + 4 = x2 + 6x
=> 4x – 6x + 4 = 0
=> -2x + 4 = 0
=> -2x = – 4
x = -4/-2
x = 2
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