Numbers and Sequences Ex 2.7-10th Std Maths-Book Back Question And Answer
Question 1.
Which of the following sequences are in G.P?
(i) 3,9,27,81,…
(ii) 4,44,444,4444,…
(iii) 0.5,0.05,0.005,
(iv) 13,16,112, ………….
(v) 1, -5, 25,-125,…
(vi) 120, 60, 30, 18,…
(vii) 16, 4, 1, 14, ……….
Answer:
Question 2.
Write the first three terms of the G.P. whose first term and the common ratio are given below.
(i) a = 6, r = 3
Answer:
a = 6, r = 3
ar = 6 × 3 = 18,
ar2 = 6 × 9 = 54
The three terms are 6, 18 and 54
(ii) a = 2–√, r = 2–√.
Answer:
ar = 2–√ × 2–√ = 2,
ar2 = 2–√ × 2 = 2 2–√
The three terms are 2–√, 2 and 22–√
(iii) a = 1000, r = 25
Answer:
ar = 1000 × 25 = 400,
ar2 = 1000 × 425 = 40 × 4 = 160
The three terms are 1000,400 and 160.
Question 3.
In a G.P. 729, 243, 81,… find t7.
Answer:
The G.P. is 729, 243, 81,….
Question 4.
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression
Solution:
G.P = x + 6, x + 12, x + 15
In G.P r = t2t1=t3t2
x+12x+6=x+15x+12
(x + 12)2 = (x + 6) (x + 5)
x2 + 24x + 144 = x2 + 6x + 15x + 90
24x – 21x = 90 – 144
3x = -54
x = −543 = -18
x = -18
Question 5.
Find the number of terms in the following G.P.
(i) 4,8,16,…,8192?
Answer
Here a = 4; r = 84 = 2
tn = 8192
a . rn-1 = 8192 ⇒ 4 × 2n-1 = 8192
2n-1 = 81924 = 2048
2n-1 = 211 ⇒ n – 1 = 11
n = 11 + 1 ⇒ n = 12
Number of terms = 12
(ii) 13, 19, 127, ……………, 12187
Answer
a = 13 ; r = 19 ÷ 13 = 19 × 31 = 13
tn = 12187
a. rn-1 = 12187
n – 1 = 6 ⇒ n = 6 + 1 = 7
Number of terms = 7
Question 6.
In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term.
Answer
Given, 9th term = 32805
a. rn-1 = 12187
t9 = 32805 [tn = arn-1]
a.r8 = 32805 …..(1)
6th term = 1215
a.r5 = 1215 …..(2)
Divide (1) by (2)
ar8ar5 = 328051215 ⇒ r3 = 6561243
= 218781 = 72927 = 2439 = 813
r3 = 27 ⇒ r3 = 33
r = 3
Substitute the value of r = 3 in (2)
a. 35 = 1215
a × 243 = 1215
a = 1215243 = 5
Here a = 5, r = 3, n = 12
t12 = 5 × 3(12-1)
= 5 × 311
∴ 12th term of a G.P. = 5 × 311
Question 7.
Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.
Solution
t8 = 768 = ar7
r = 2
t10 = ar9 = ar7 × r × r
= 768 × 2 × 2 = 3072
Question 8.
If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.
Answer:
a, b, c are in A.P.
t2 – t1 = t3 – t2
b – a = c – b
2b = a + c …..(1)
3a, 3b, 3c are in G.P.
From (1) and (2) we get
3a, 3b, 3c are in G.P.
Question 9.
In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 572. Find the three terms.
Answer:
Let the three terms of the G.P. be ar, a, ar
Product of three terms = 27
ar × a × ar = 27
a3 = 27 ⇒ a3 = 33
a = 3
Sum of the product of two terms taken at a time is 57/2
6r2 – 13r + 6 = 0
6r2 – 9r – 4r + 6 = 0
3r (2r – 3) -2(2r – 3) = 0
(2r – 3) (3r – 2) = 0
2r – 3 = 0 or 3r – 2 = 0
2r = 3 (or) 3r – 2 = 0
r = 32 (or) r = 23
∴ The three terms are 2, 3 and 92 or 92, 3 and 2
Question 10.
A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
Answer:
Starting salary (a) = ₹ 60000
Increased salary = 5% of starting salary
= 5100 × 60000
= ₹ 3000
Starting salary for the 2nd year = 60000 + 3000
= ₹ 63000
Year increase = 5% of 63000
= 5100 × 63000
= ₹ 3150
Starting salary for the 3rd year = 63000 + 3150
= ₹ 66150
60000, 63000, 66150,…. form a G.P.
a = 60000; r = 6300060000 = 6360 = 2120
tn = ann-1
t5 = (60000) (2120)4
= 60000 × 2120 × 2120 × 2120 × 2120
= 6×21×21×21×212×2×2×2
= 72930.38
5% increase = 5100 × 72930.38
= ₹ 3646.51
Salary after 5 years = ₹ 72930.38 + 3646.51
= ₹ 76576.90
= ₹ 76577
Question 11.
Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.
Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
What is his salary in the 4th year with respect to the offers A and B?
Answer::
Starting salary (a) = ₹ 20,000
Annual increase = 6% of 20000
= 5100 × 20000
= ₹ 1200
Salary for the 2nd year = ₹ 20000 + 1200
= ₹ 21200
Here a = 20,000; r = 2120020000 = 212200 = 106100 = 5350
n = 4 years
tn = arn-1
Salary at the end of 4th year = 23820
For B
Starting salary = ₹ 22000
(a) = 22000
Annual increase = 3% of 22000
= 3100 × 22000
= ₹ 660
Salary for the 2nd year = ₹ 22000 + ₹ 660
= ₹ 22,660
Here a = 22000; r = 2266022000
= 22662200 = 11331100 = 103100
Salary at the end of 4th year = 22000 × (103100)4-1
= 22000 × (103100)3
= 22000 × 103100 × 103100 × 103100
= 24039.99 = 24040
4th year Salary for A = ₹ 23820 and 4th year Salary for B = ₹ 24040
Question 12.
If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb-c × yc-a × za-b = 1
Answer:
a, b, c are three consecutive terms of an A.P
∴ a = a, b = a + dand c = a + 2d respectively ….(1)
x, y, z are three consecutive terms of a G.P
∴ x = x, y = xr, z = xr2 respective ……(2)
L.H.S = xb-c × yc-a × za-b ( Substitute the values from 1 and 2 we get)
L.H.S = R.H.S
Hence it is proved
