Inter 1st Year Maths Sequences and Series Solutions Exercise 8b

Practicing the AP Board Solutions Class 11 Maths and Chapter 8 Inter 1st Year Maths Sequences and Series Solutions Exercise 8b Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Sequences and Series Solutions Exercise 8b

Sequences and Series Exercise 8b Solutions

Sequences and Series Class 11 Exercise 8b Solutions – Sequences and Series 8b Exercise Solutions

I.

Question 1.
Find the 20th and nth terms of the G.P. \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots \ldots \ldots\)
Solution:

Question 2.
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution:
Given the common ratio of the G.P., r = 2
And, let a be the first term of the G.P.
Now,

Question 3.
The 5th, 8th, and 11th terms of a G.P. are p, q, and s, respectively. Show that q² = ps.
Solution:
Let’s take a to be the first term and r to be the common ratio of the G.P.
Then, according to the question, we have
a₅ = ar^5-1 = ar⁴ = p ……..(i)
a₈ = ar^8-1 = ar⁷ = q ………(ii)
a₁₁ = ar^11-1 = ar¹⁰ = s ………..(iii)
Dividing equation (ii) by (i) we get,
\(\frac{a^7}{a r^4}=\frac{q}{p}\)
r³ = \(\frac {q}{p}\) ………(iv)
On dividing equation (iii) by (ii), we get,
\(\frac{a r^10}{a r^7}=\frac{s}{q}\)
r³ = \(\frac {s}{q}\) ………(v)
Equating the values of r³ obtained in (iv) and (v), we get
\(\frac{q}{p}=\frac{s}{q}\)
q² = ps
Hence proved.

Question 4.
The 4th term of a G.P. is the square of its second term, and the first term is -3. Determine its 7th term.
Solution:
Let’s consider a to be the first term and r to be the common ratio of the G.P.
Given a = -3
and we know that,
a_n = ar^n-1
so, a₄ = ar³ = (-3) r³
a₂ = ar¹ = (-3) r
Then, from the question, we have
(-3) r³ = [(-3) r]²
⇒ -3r³ = 9r²
⇒ r = -3
∴ a₇ = ar^7-1
= ar⁶
= (-3) (-3)⁶
= (3)⁷
= -2187
Therefore, the seventh term of the G.P. is -2187.

Question 5.
Which term of the following sequences:
(a) 2, 2√2, 4,….. is 128?
(b) √3, 3, 3√3, …….. is 729?
(c) \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots \text { is } \frac{1}{19683}\)?
Solution:
(a) The given sequence, 2, 2√2, 4,……….
we have, a = 2 and r = 2√2 ÷ 2 = √2
Taking the nth term of this sequence as 128, we have

Therefore, the 13th term of the given sequence is 128.

(b) Given the sequence, √3, 3, 3√3,…….
we have, a = √3 and r = 3 ÷ √3 = √3
Taking the nth term of this sequence to be 729, we have

Equating the exponents, we have
\(\frac{1}{2}+\frac{n-1}{2}\) = 6
\(\frac{1+n-1}{2}\) = 6
∴ n = 12
Therefore, the 12th term of the given sequence is 729.

(c) Given sequence, \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}\),…….
a = \(\frac {1}{3}\) and r = \(\left(\frac{1}{9}\right)\left(\frac{1}{3}\right)=\left(\frac{1}{3}\right)\)
Taking the nth term of this sequence to be \(\frac {1}{19683}\), we have

Therefore, the 9th term of the given sequence is \(\frac {1}{19683}\).

Question 6.
For what values of x, the numbers \(-\frac{2}{7}, x,-\frac{7}{2}\) are in G.P?
Solution:
The given numbers are \(-\frac{2}{7}, x,-\frac{7}{2}\)
common ratio = \(\frac{x}{\left(\frac{-2}{7}\right)}=\frac{-7 x}{2}\)
Also, common ratio = \(\frac{\left(\frac{-7}{2}\right)}{x}=\frac{-7}{2} x\)
∴ \(\frac{-7 x}{2}=\frac{-7}{2 x}\)
⇒ x² = \(\frac{-2 \times 7}{-2 \times 7}\)
⇒ x = 1
⇒ x = √1
⇒ x = ±1
Therefore, for x = ±1, the given numbers will be in G.P.

II. Find the sum of the indicated number of terms in each of the geometric progressions in Exercises 1 to 4:

Question 1.
0.15, 0.015, 0.0015,…….. 20 terms.
Solution:
Given G.P., 0.15, 0.015, 0.00015,………
Here, a = 0.15 and r = \(\frac{0.015}{0.15}\) = 0.1
We know that, Sn = \(\frac{a\left(1-r^n\right)}{(1-r)}\)

Question 2.
√7, √21, 3√7,……… n terms.
Solution:
The given G.P. is √7, √21, 3√7,………
Here, a = √7 and

Question 3.
1, -a, a², -a³,… n terms (if a ≠ -1).
Solution:
The given G.P is 1, -a, a², -a³,…..
Here, the first term = a₁ = 1
And the common ratio = r = -a
We know that,

Question 4.
x³, x⁵, x⁷,…… n terms (if x ≠ ±1).
Solution:
Given G.P is x³, x⁵, x⁷,………
Here, we have a = x³ and r = \(\frac{x^5}{x^3}\) = x²
We know that, S_n = \(\frac{a\left(1-r^n\right)}{(1-r)}\)
= \(\frac{x^3\left[1-\left(x^2\right)^n\right]}{1-x^2}\)
= \(\frac{x^3\left(1-x^{2 n}\right)}{1-x^2}\)

Question 5.
Evaluate \(\sum_{k=1}^n\left(2+3^k\right)\).
Solution:

We can see that the terms of this sequence, 3, 3², 3³,… form a G.P.
And, we know

On substituting the above value in equation (1), we get
\(\sum_{k=1}^{11}\left(2+3^k\right)=22+\frac{3}{2}\left(3^{11}-1\right)\)

Question 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.
Solution:
Let \(\frac {a}{r}\), a, ar be the first three terms of the G.P.
\(\frac {a}{r}\) + a + ar = \(\frac {39}{10}\) …….(1)
(\(\frac {a}{r}\)) (a) (ar) = 1 …..(2)
from (2), we have a³ = 1
Hence, a = 1 [considering real roots only]
Substituting the value of a in (1), we get
\(\frac{1}{r}+1+r=\frac{39}{10}\)
⇒ \(\frac{\left(1+r+r^2\right)}{r}=\frac{39}{10}\)
⇒ 10 + 10r + 10r² = 39r
⇒ 10r² – 29r + 10 = 0
⇒ 10r² – 25r – 4r + 10 = 0
⇒ 5r(2r – 5) – 2(2r – 5) = 0
⇒ (5r – 2) (2r – 5) = 0
Thus, r = \(\frac {2}{5}\) or \(\frac {5}{2}\)
Therefore, the three terms of the G.P. are \(\frac {5}{2}\), 1, and \(\frac {2}{5}\).

Question 7.
How many terms of G.P. 3, 3², 3³,… are needed to give the sum 120?
Solution:
Given G.P. is 3, 3², 3³,………
Let’s consider that n terms of this G.P. are required to obtain the sum 120.
We know that S_n = \(\frac{a\left(r^n-1\right)}{(r-1)}\)
Here a = 3 and r = 3
S_n = 120 = \(\frac{3\left(3^{\mathrm{n}}-1\right)}{3-1}\)
⇒ 120 = \(\frac{3\left(3^n-1\right)}{2}\)
⇒ \(\frac{120 \times 2}{3}\) = 3n – 1
⇒ 3n – 1 = 80
⇒ 3n = 81
⇒ 3n = 34
Equating the exponents, we get n = 4
Therefore, four terms of the given G.P. are required to obtain the sum 120.

Question 8.
The sum of the first three terms of a G.P. is 16, and the sum of the next three terms is 128. Determine the first term, the common ratio, and the sum to n terms of the G.P.
Solution:
Let’s assume the G.P. to be a, ar, ar², ar³,……….
Then, according to the question, we have
a + ar + ar² = 16 and ar³ + ar⁴ + ar⁵ = 128
a(1 + r + r²) = 16 ……..(1)
and ar³ (1 + r + r²) = 128 ……..(2)
Dividing equation (2) by (1), we get
\(\frac{a^3\left(1+r+r^2\right)}{a(1+r+r)}\) = \(\frac {128}{16}\)
⇒ r³ = 8
⇒ r = 2
Now, using r = 2 in (1), we get
a(1 + 2 + 4) = 16
⇒ a(7) = 16
⇒ a = \(\frac {16}{7}\)
Now, the sum of terms is given as
S_n = \(\frac{a\left(r^n-1\right)}{(r-1)}\)
⇒ S_n = \(\frac{16}{7} \frac{\left(2^n-1\right)}{2-1}=\frac{16}{7}\left(2^n-1\right)\)

Question 9.
Given a G.P. with a = 729 and 7th term 64, determine S₇.
Solution:
Given, a = 729 and a₇ = 64
Let r be the common ratio of the G.P.
Then, we know that a_n = ar^n-1

And we know that

Question 10.
Find a G.P. for which the sum of the first two terms is -4, and the fifth term is 4 times the third term.
Solution:
Consider a to be the first term and r to be the common ratio of the G.P.
Given, S₂ = -4
Then, from the question, we have

Using the value of r in (1), we have

Therefore, the required G.P. is \(\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \ldots \ldots\) or 4, -8, 16, -32,………

Question 11.
If the 4th, 10th, and 16th terms of a G.P. are x, y, and z, respectively, then prove that x, y, and z are in G.P.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a₄ = ar³ = x ……….(1)

Therefore, x, y, z are in G.P.

Question 12.
Find the sum of n terms of the sequence, 8, 88, 888, 8888,………
Solution:
Given sequence 8, 88, 888, 8888,……..
This sequence is not a G.P.
But it can be changed to G.P. by writing the terms as
S_n = 8 + 88 + 888 + 8888 + ……… to n terms.
= \(\frac {8}{9}\) [9 + 99 + 999 + 9999 + ……… to n terms]
= \(\frac {8}{9}\) [(10 – 9) + (10² – 1) + (10³ – 1) + (10⁴ – 1) + …….. to n terms]

Question 13.
Find the sum of the products of the corresponding terms of the two sequences 2, 4, 8, 16, 32, and 128, 32, 8, 2, \(\frac {1}{2}\).
Solution:
The required sum = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 × \(\frac {1}{2}\)
= 64 [4 + 2 + 1 + \(\frac {1}{2}\) + \(\frac{1}{2^2}\)]
Now, it’s seen that 4, 2, 1, \(\frac{1}{2}\), \(\frac{1}{2^2}\) form a G.P.
with the first term a = 4
common ratio, r = \(\frac {1}{2}\)
we know

Therefore, the required sum = \(64\left(\frac{31}{4}\right)\)
= (16) (31)
= 496

Question 14.
Show that the products of the corresponding terms of the sequences a, ar, ar²,…, ar^n-1 and A, AR, AR²,…, AR^n-1 form a G.P., and find the common ratio.
Solution:
To prove the sequence, aA, arAR, ar²AR²,…, ar^n-1AR^n-1, forms a G.P.
Now, we have
\(\frac{\text { Second term }}{\text { First term }}=\frac{\mathrm{arAR}}{\mathrm{aR}}\) = rR
\(\frac{\text { Third term }}{\text { Second term }}=\frac{\mathrm{ar}^2 \mathrm{AR}^2}{{arAR}}\) = rR
Therefore, the above sequence forms a G.P. and the common ratio is rR.

III.

Question 1.
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Solution:
Consider a to be the first term and r to be the common ratio of the G.P.
Then, a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³
From the question, we have
a₃ = a₁ + 9
ar² = a + 9 ………(i)
a₂ = a₄ + 18
ar = ar³ + 18 ……..(ii)
So, from (i) and (ii), we get
a(r² – 1) = 9 ……..(iii)
ar(1 – r²) = 18 ……….(iv)
Now dividing (iv) by (iii), we get
\(\frac{{ar}\left(1-r^2\right)}{a\left(r^2-1\right)}=\frac{18}{9}\)
⇒ -r = 2
⇒ r = -2
On substituting the value of r in (i), we get
4a = a + 9
⇒ 3a = 9
⇒ a = 3
Therefore, the first four numbers of the G.P. are 3, 3(-2), 3(-2)2 and 3(-2)3
i.e., 3, -6, 12, and -24.

Question 2.
If the pth, qth, and rth terms of a G.P. are a, b, and c, respectively. Prove that, a^q-r b^r-p c^p-q = 1
Solution:
Let’s take A to be the first term and R to be the common ratio of the G.P.
Then, according to the question, we have

Hence proved.

Question 3.
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.
Solution:
Given that the first term of the G.P. is a, and the last term is b.
Thus, the G.P. is a, ar, ar², ar³,……., ar^n-1
where r is the common ratio.
Then, b = ar^n-1 ……..(1)
P = Product of n terms

And, the product of n terms p is given by,
P = \(a^n r^{\frac{n(n-1)}{2}}\)

Question 4.
Show that the ratio of the sum of the first n terms of a G.P. to the sum of terms from the (n + 1)th to (2n)th term is \(\frac{1}{\mathrm{r}^{\mathrm{n}}}\)
Solution:
Let a be the first term and r be the common ratio of the G.P.
Sum of first n terms = \(\frac{a\left(1-r^n\right)}{(1-r)}\)
Since there are n terms from (n + 1)th to (2n)th term,
Sum of terms from (n + 1)th to (2n)th term = \(\frac{a_{n+1}\left(1-r^n\right)}{(1-r)}\)
\(a^{n+1}=a r^{n+1-1}=a r^n\)
Thus, the required ratio = \(\frac{a\left(1-r^n\right)}{(1-r)} \times \frac{(1-r)}{a r^n\left(1-r^n\right)}=\frac{1}{r^n}\)
Thus, the ratio of the sum of the first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is \(\frac{1}{\mathrm{r}^{\mathrm{n}}}\).

Question 5.
If a, b, c, and d are in G.P., show that (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)².
Solution:
Given that a, b, c, and d are in G.P.
So, we have.
bc = ad ………(1)
b² = ac ……….(2)
c² = bd ………..(3)
Taking the R.H.S., we have
R.H.S = (ab + bc + cd)²
= (ab + ad + cd)² [using (1)]
= [ab + d(a + c)]²
= a²b² + 2abd (a + c) + d² (a + c)²
= a²b² + 2a²c²+ 2b²c² + d²d² + 2d²b² + d²c² [using (1) and (2)]
= a²b² + a²c² + a²c² + b²c² + b²c² + d²a² + d²b² + d²b² + d²c²
= a²b² + a²c² + a²d² + b²b² + b²c² + b²d² + c²b² + c²c² + c²d² [Using (2) and (3) and rearranging terms]
= a² (b² + c² + d²) + b² (b² + c² + d²) + c² (b² + c² + d²)
= (a² + b² + c²) (b² + c² + d²)
= L.H.S.
Thus, L.H.S. = R.H.S.
Therefore, (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

Question 6.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution:
Let’s assume G₁ and G₂ to be two numbers between 3 and 81 such that the series 3, G₁, G₂, 81 forms a G.P.
and let a be the first term and r be the common ratio of the G.P.
Now, we have the 1st term as 3 and the 4th term as 81.
81 = (3) (r)³
⇒ r³ = 27
⇒ r = 3 (taking real roots only)
For r = 3,
G₁ = ar = (3) (3) = 9
G₂ = ar² = (3) (3)² = 27
Therefore, the two numbers that can be inserted between 3 and 81 so that the resulting sequence becomes a G.P. are 9 and 27.

Question 7.
Find the value of n so that \(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\) may be the geometric mean between a and b.
Solution:
We know that,
The G.M. of a and b is given by \(\sqrt{a b}\)
Then, from the question, we have
\(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{a b}\)
By squaring both sides, we get

2n + 1 = 0 (Equating the exponents)
∴ n = \(\frac {-1}{2}\)

Question 8.
The sum of two numbers is 6 times their geometric mean. Show that numbers are in the ratio (3 + 2√2) : (3 – 2√2).
Solution:
Consider the two numbers to be a and b.
The G.M. = √ab
From the question, we have
a + b = 6√ab …………(1)
⇒ (a + b)² = 36(ab)
also, (a – b)² = (a + b)² – 4ab
⇒ (a + b)² = 36ab – 4ab
⇒ (a + b)² = 32ab
⇒ a – b = √32 √ab
⇒ a – b = 4√2 √ab ……….(2)
On adding (1) and (2), we get
2a = (6 + 4√2)√ab
⇒ a = (3 + 2√2)√ab
Substituting the value of a in (1), we get
b = 6√ab – (3 + 2√2)√ab
b = (3 – 2√2)√ab
\(\frac{a}{b}=\frac{(3+2 \sqrt{2}) \sqrt{a b}}{(3-2 \sqrt{2}) \sqrt{a b}}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}\)
Therefore, the required ratio is (3 + 2√2) : (3 – 2√2)

Question 9.
If A and G are A.M. and G.M., respectively, between two positive numbers, prove that the numbers are A ± \(\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\).
Solution:
Given that A and G are A.M. and G.M. between two positive numbers.
And, let these two positive numbers be a and b.
So, AM = A = \(\frac{a+b}{2}\) …..(1)
GM = G = √ab ………(2)
From (1) and (2), we get
a + b = 2A ………(3)
ab = G² ………..(4)
substituting the value of a and b from (3) and (4) in the identity (a-b)² = (a + b)² – 4ab,
we have (a – b)² = 4A² – 4G² = 4(A² – G²)
(a – b)² = 4(A + G) (A – G)
(a – b) = 2\(\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\) ………..(5)
From (3) and (5), we get
2a = 2A + 2\(\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\)
⇒ a = A + \(\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\)
Substituting the value of a in (3), we have
b = 2A – A
= A
= \(\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\)
Therefore, the two numbers are A ± \(\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\)

Question 10.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour?
Solution:
Given that the number of bacteria doubles every hour.
Hence, the number of bacteria after every hour will form a G.P
Here we have, a = 30 and r = 2
so, a₃ = ar²
= (30) (2)²
= 120
Thus, the number of bacteria at the end of the 2nd hour will be 120.
and, a₅ = ar⁴
= (30) (2)⁴
= 480
The number of bacteria at the end of the 4th hour will be 480.
a_n+1 = arⁿ = (30) 2ⁿ
Therefore, the number of bacteria at the end of the nth hour will be 30(2)ⁿ.

Question 11.
What will Rs. 500 amount to in 10 years after its deposit in a bank that pays an annual interest rate of 10% compounded annually?
Solution:
Given that the amount deposited in the bank is Rs. 500.
At the end of first year, amount = Rs. 500 \(\left(1+\frac{1}{10}\right)\) = Rs. 500 (1.1)
At the end of the 2nd year, amount = Rs. 500 (1.1) (1.1)
At the end of the 3rd year, amount = Rs. 500 (1.1) (1.1) (1.1) and so on
Therefore, the amount at the end of 10 years = Rs. 500 (1.1) (1.1)………. (10 times) = Rs. 500 (1.1)¹⁰

Question 12.
If A.M. and G.M. of the roots of a quadratic equation are 8 and 5, respectively. Then obtain the quadratic equation.
Solution:
Let’s consider the roots of the quadratic equation to be a and b.
Then, we have
A.M = \(\frac{a+b}{2}\) = 8
⇒ a + b = 16 ………(1)
G.M = √ab = 5
⇒ ab = 25 ………..(2)
We know that A quadratic equation can be formed as,
x² – x(sum of roots) + (Product of roots) = 0
⇒ x² – x² (a + b) + (ab) = 0
⇒ x² – 16x + 25 = 0 [using (1) and (2)]
Therefore, the required quadratic equation is x² – 16x + 25 = 0.