Class 9 Advanced Maths Chapter 6 Exploring Some More Progressions Ex 6.1 Solutions - #NCSOLVE 📚

Accurate and updated NCERT Class 9 Advanced Maths Solutions Chapter 6 Exploring Some More Progressions Ex 6.1 help students build a strong mathematical foundation.

Ex 6.1 Class 9 Advanced Maths Solutions

Advanced Maths Class 9 Exercise 6.1 Solutions

Exercise 6.1 Class 9 Advanced Maths Solutions

Question 1.
Find the sum of the series 0.15 + 0.015 + 0.0015 +… to 15 terms.
Solution:
Given series is 0.15 + 0.015 + 0.0015+… .
We observe that the given series is a GP.
Here, a = 0.15 and r = \(\frac{0.015}{0.15}\) = 0.01
Now, the sum of first n terms of a GP is given by
S_n = a(\(\frac{1-r^n}{1-r}\)).
On substituting all the values, we get
S₁₅ = 0.15(\(\frac{1-(0.1)^{15}}{1-0.1}\)) = 0.15(\(\frac{1-10^{-15}}{0.9}\))
Therefore, S₁₅ = \(\frac{1}{6}\) (1 – 10¹⁵)

Question 2.
Find the sum to n terms of the series 0.9 + 0.99 + 0.999 +… [Hint Write (0.9 = 1 – 0.1) and (0.99 = 1 – 0.01) and so on]
Solution:
Given series is 0.9 + 0.99 + 0.999 +….
Let S_n = 0.9 + 0.99 + 0.999 + … upto n terms.
Now, the sum to n terms becomes
S_n = (1 – 0.1) + (1 – 0.01) + (1 – 0.001) + … upto n terms

Hence, S_n = n – (0.1 + 0.01 + 0.001 +… upto n terms)
Now, the series 0.1 + 0.01 + 0.001 +… is a GP with a = 0.1 and r = 0.1.
We know that the sum of first n terms of a GP is
S_n = a(\(\frac{1-r^n}{1-r}\)).
So, 0.1 + 0.01 + 0.001 + … = \(\frac{0.1\left(1-0.1^n\right)}{1-0.1}\)
= \(\frac{0.1\left(1-0.1^n\right)}{0.9}\) = \(\frac{1-(0.1)^n}{9}\)
Therefore, S_n = n – \(\frac{1-\left(10^{-1}\right)^n}{9}\)
Hence, the sum of n terms of the given series is
S_n = n – \(\frac{1-10^{-n}}{9} .\)

Question 3.
Find the sum to n terms of the series 5 + 55 + 555 +… .
Solution:
Given series is 5 + 55 + 555 +… .
Let the sum to n terms be S_n.
∴ S_n = \(\frac{5}{9}\)(9 + 99 + 999 + … upto to n terms)
= \(\frac{5}{9}\)[(10 – 1) +(10² – 1) + (10³ – 1) + … upto n terms]
= \(\frac{5}{9}\)[(10 + 10² + 10³ + … to n terms) – (1 + 1 + 1 +… upto n terms)]
= \(\frac{5}{9}\)(10 + 10² +… + 10ⁿ – n) ……. (i)
Also, the series 10 + 10² +… + 10ⁿ is a GP with a = 10 and r = 10.
We know that the sum of first n terms of a GP
S_n = a(\(\frac{r^n-1}{r-1}\))
So, 10 + 10² + …+ 10ⁿ = \(\frac{10\left(10^n-1\right)}{9}\)
On substituting this value in Eq. (i), we get
S_n = \(\frac{5}{9}\)[\(\frac{10\left(10^n-1\right)}{9}\) – n] = \(\frac{50}{81}\)(10ⁿ – 1) – \(\frac{5 n}{9}\)

Question 4.
The sum of the First n terms of the sequence 3, 6,12, … is 381. Find n.
Solution:
Given sequence is 3,6,12, …
We observe that the given sequence is a GP with
a = 3 and r = \(\frac{6}{3}\) = 2.
Also, it is given that S_n = 381
Now, the sum of first n terms of a GP is
S_n = a(\(\frac{r^n-1}{r-1}\))
On substituting all the values, we get
381 = 3(\(\frac{2^n-1}{2-1}\)) ⇒ 381 = 3(2ⁿ – 1)
On dividing both sides by 3, we get
127 = 2ⁿ – 1 ⇒ 2ⁿ = 127 + 1 ⇒ 2ⁿ = 128
Now, 128 = 2⁷
Therefore, n = 7
Hence, the required value of n is 7.

Question 5.
Find the sum to n terms of the series
x(x + y) + x² (x² + y²) + x³ (x³ + y³) + … .
Solution:
Given series is
x(x + y) + x² (x² + y²) + x³ (x³ + y³) + … .
Let S_n = x(x + y) + x² (x² + y²) + x³ (x³ + y³) + … upto n terms
Now, x(x + y) = x² + xy
x²(x² + y²) = x⁴ + x²y²
x³(x³ + y³) = x⁶ + x³y³ and so on.
Therefore, the series becomes
S_n = (x² + x⁴ + x⁶ + …upto n terms) + (xy + x²y² + x³y³ +… upto n terms)
Now, both are GPs.
For the first GP, a = x², and r = x²
We know that S_n = a(\(\frac{1-r^n}{1-r}\))
∴ x² + x⁴ + x⁶ + … = \(\frac{x^2\left(1-x^{2 n}\right)}{1-x^2}\)
Also, for the second GP, a = xy, and r = xy
Therefore, xy + x²y² + x³y³ +… = \(\frac{x y\left(1-(x y)^n\right)}{1-x y}\)
On adding both sums, we get
S_n = \(\frac{x^2\left(1-x^{2 n}\right)}{1-x^2}\) + \(\frac{x y\left(1-(x y)^n\right)}{1-x y}\)
Hence, the sum to ti terms of the given series is
S_n = \(\frac{x^2\left(1-x^{2 n}\right)}{1-x^2}\) + \(\frac{x y\left(1-(x y)^n\right)}{1-x y}\)

Question 6.
Find the sum of the infinite terms of the series
1 + \(\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^2}+\frac{1}{3^4}+\frac{1}{2^3}+\frac{1}{3^6}+\ldots .\)
Solution:

Question 7.
The sum of an infinite series of GP is 6 and sum of the squares of these terms is 12. Find the common ratio of the original GP.
Solution:
Let the first term of the GP be a and common ratio be r.
It is given that the sum of the infinite GP is 6.
We know that S_∞ = \(\frac{a}{1-r}\)
∴ \(\frac{a}{1-r}\) = 6 ⇒ a = 6(1 – r) …. (i)
Now, the squares of the terms form another GP.
a², a²r², a²r⁴, ….,
whose first term is a² and common ratio is r².
Also, it is given that the sum of the squares of the terms is 12.
Therefore, \(\frac{a^2}{1-r^2}\) = 12
On substituting the value of a from Eq. (i), we get
\(\frac{[6(1-r)]^2}{1-r^2}\) = 12
⇒ \(\frac{36(1-r)^2}{(1-r)(1+r)}\) = 12 ⇒ \(\frac{36(1-r)}{1+r}\) = 12
On dividing both sides by 12, we get
\(\frac{3(1-r)}{1+r}\) = 1
⇒ 3 – 3r = 1 + r ⇒ 3 – 1 = 4r ⇒ 2 = 4r
⇒ r = \(\frac{1}{2}\)
Hence, the common ratio of the original GP is r = \(\frac{1}{2}\).

Question 8.
If the sum to infinity of the series 1 + r + r² + r³ +… is S, given that |r| < 1 then write r in terms of S.
Solution:
Given, 1 + r + r² + r³ +… is an infinite GP with first term (a) = 1 and common ratio = r.
Also, it is given that the sum of infinity of the series is S.
Now, for an infinite GP with |r| < 1,
S_∞ = \(\frac{a}{1-r}\)
On substituting a = 1, we get S = \(\frac{1}{1-r}\)
Therefore, 1 – r = \(\frac{1}{S}\)
⇒ r = 1 – \(\frac{1}{S}\)
Hence, r = \(\frac{S – 1}{S}\)

Question 9.
Find the value of \(4^{\frac{1}{2}} \cdot 4^{\frac{1}{4}} \cdot 4^{\frac{1}{8}} \cdot 4^{\frac{1}{16}} \ldots \text { to } \infty .\)
Solution:
Given expression is
\(4^{\frac{1}{2}} \cdot 4^{\frac{1}{4}} \cdot 4^{\frac{1}{8}} \cdot 4^{\frac{1}{16}} \ldots \text { to } \infty .\)
Using the law of exponents,
a^m.aⁿ = a^{m + n}

Question 10.
The mid-points D, E and F of the sides of an equilateral ΔABC are joined to form another smaller equilateral triangle. This process is repeated in the ΔDEF and so on indefinitely, getting smaller and smaller triangle. If each side of the ΔABC is 16 cm, find the sum of the perimeters of all the triangles, so formed.

Solution:
Given, the side of the equilateral ΔABC is 16 cm.
Therefore, perimeter of ΔABC is P₁ = 3 × 16 = 48 cm.
Now, the triangle formed by joining the mid-points of the sides has each side equal to half of the previous triangle.
Hence, side of the next triangle is \(\frac{16}{2}\) = 8 cm.
Therefore, its perimeter is P₂ = 3 × 8 = 24 cm
Similarly, the next perimeters are 12 cm, 6 cm, … .
Thus, the perimeters form an infinite GP
48 + 24 + 12 + 6 + … .
Here, a = 48, r = \(\frac{24}{48}\) = \(\frac{1}{2}\)
Since, r < 1, the sum of the infinite GP is
S_∞ = \(\frac{a}{1 – r}\).
Therefore, S_∞ = \(\frac{48}{1-\frac{1}{2}}=\frac{48}{\frac{1}{2}}\) = 96
Hence, the sum of the perimeters of all the triangles, so formed is 96 cm.

Question 11.
Let f(x) = 2x + 1 then find the number of values of x for which f(x), f(2x), f(4x) are in a GP.
Solution:
Given, f(x) = 2x + 1
∴ f(2x) = 2(2x) + 1 = 4x + 1
and f(4x) = 2(4x) + 1 = 8x + 1
Now, the three numbers 2x + 1, 4x + 1, 8x + 1 are in GP.
For the three numbers a, b, c to be in GP,
b² = ac.
Therefore, (4x + 1)² = (2x + 1) (8x + 1)
On expanding both sides, we get
16x² + 8x + 1 = 16x² + 10x + 1
⇒ 8x = 10x
⇒ 2x = 0
⇒ x = 0
Hence, there is only one value of x.

Question 12.
If t₁, t₂, t₃, t₄,… are the terms of a GP, whose common ratio is r such that -1 < r< 1 then evaluate ‘ the following
\(\frac{t_1-t_3+t_5-\ldots}{t_2-t_4+t_6-\ldots}\)
Solution:
Let the first term of the GP be a and common ratio be r, where -1 < r < 1.
Then, t₁ = a, t₂ = ar, t₃ = ar², t₄ = ar³, …
Now, t₁ – t₃ + t₅ – … = a – ar² + ar⁴ -,…
⇒ t₁ – t₃ + t₅ – … = a(1 – r² + r⁴ – ……. (i)
This is an infinite GP with a₁ = 1, r₁ = – r².
We know that S_∞ = \(\frac{a}{1 – r}\).

Hence, the required value is \(\frac{1}{r}\)..

Question 13.
A particular ball rebounds (\(\frac{3}{5}\))th of the height from which it falls, whenever it strikes the floor. Find the total distance it covers before coming to rest, if it falls due to gravity from a height of 90 m.
Solution:
The ball is dropped from a height of 90 m
and each time it rebounds to (\(\frac{3}{5}\))th of the previous height.
Therefore, first rebound height is 90 × \(\frac{3}{5}\) = 54 m.
and second rebound height is 54 × \(\frac{3}{5}\) and so on.
Now, the total distance travelled by the ball is
90 + 2(54 + 54.\(\frac{3}{5}\) + 54.(\(\frac{3}{5}\))² + …).
Here, 54 + 54.\(\frac{3}{5}\) + 54.\(\frac{3}{5}\))² + …
is an infinite GP with a = 54 and r = \(\frac{3}{5}\).
Since, |r| < 1, the sum of the infinite GP is a
S_∞ = \(\frac{a}{1 – r}\)
Therefore, S_∞ = \(\frac{54}{1-\frac{3}{5}}=\frac{54}{\frac{2}{5}}\)
54 × \(\frac{5}{2}\) = 135
Hence, total distance travelled = 90 + 2(135)
= 90 + 270
= 360
Hence, the total distance covered by the ball before coming to rest is 360 m.

The post Class 9 Advanced Maths Chapter 6 Exploring Some More Progressions Ex 6.1 Solutions appeared first on Learn CBSE.

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