Exercise 8.1 Solutions

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

Q1. a_n = n (n + 2)
Solution:
Given,

n^th term of a sequence a_n = n (n + 2) 

On substituting n = 1, 2, 3, 4, and 5, we get the first five terms

a₁ = 1(1 + 2) = 3

a₂ = 2(2 + 2) = 8

a₃ = 3(3 + 2) = 15

a₄ = 4(4 + 2) = 24

a₅ = 5(5 + 2) = 35

Hence, the required terms are 3, 8, 15, 24, and 35.

Q2. a_n = n/n+1
Solution:
Given the n^th term, a_n = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6.

Q3. a_n = 2ⁿ
Solution:
Given the n^th term, a_n = 2ⁿ

On substituting n = 1, 2, 3, 4, 5, we get

a₁ = 2¹ = 2

a₂ = 2² = 4

a₃ = 2³ = 8

a₄ = 2⁴ = 16

a₅ = 2⁵ = 32

Hence, the required terms are 2, 4, 8, 16, and 32.

Q4.  a_n = (2n – 3)/6
Solution:
Given the n^th term, a_n = (2n – 3)/6

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6..

Q5. a_n = (-1)^n-1 5ⁿ⁺¹
Solution:
Given the n^th term, a_n = (-1)^n-1 5ⁿ⁺¹

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 25, –125, 625, –3125, and 15625.

Q6.

Solution: On substituting n = 1, 2, 3, 4, 5, we get the first 5 terms.

Hence, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n^th terms are:

Q7. a_n = 4n – 3; a₁₇, a₂₄
Solution:
Given,

The n^th term of the sequence is a_n = 4n – 3

On substituting n = 17, we get

a₁₇ = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a₂₄ = 4(24) – 3 = 96 – 3 = 93

Q8. a_n = n²/2ⁿ ; a⁷
Solution:
Given,

The n^th term of the sequence is a_n = n²/2ⁿ

Now, on substituting n = 7, we get

a₇ = 7²/2⁷ = 49/ 128

Q9. a_n = (-1)^n-1 n³; a₉
Solution:
Given,

The n^th term of the sequence is a_n = (-1)^n-1 n³

On substituting n = 9, we get

a₉ = (-1)^9-1 (9)³ = 1 x 729 = 729

Q10

Solution: On substituting n = 20, we get

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

Q11. a₁ = 3, a_n = 3a_n-1 + 2 for all n > 1
Solution:
Given, a_n = 3a_n-1 + 2 and a₁ = 3

Then,

a₂ = 3a₁ + 2 = 3(3) + 2 = 11

a₃ = 3a₂ + 2 = 3(11) + 2 = 35

a₄ = 3a₃ + 2 = 3(35) + 2 = 107

a₅ = 3a₄ + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

Q12. a₁ = -1, a_n = a_n-1/n, n ≥ 2
Solution:
Given,

a_n = a_n-1/n and a₁ = -1

Then,

a₂ = a₁/2 = -1/2

a₃ = a₂/3 = -1/6

a₄ = a₃/4 = -1/24

a₅ = a₄/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

Q13. a₁ = a₂ = 2, a_n = a_n-1 – 1, n > 2
Solution:
Given,

a₁ = a₂, a_n = a_n-1 – 1

Then,

a₃ = a₂ – 1 = 2 – 1 = 1

a₄ = a₃ – 1 = 1 – 1 = 0

a₅ = a₄ – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

Q14. The Fibonacci sequence is defined by

1 = a₁ = a₂ and a_n = a_{n – 1} + a_{n – 2}, n > 2

Find a_n+1/a_n, for n = 1, 2, 3, 4, 5
Solution:
Given,

1 = a₁ = a₂

a_n = a_{n – 1} + a_{n – 2}, n > 2

So,

a₃ = a₂ + a₁ = 1 + 1 = 2

a₄ = a₃ + a₂ = 2 + 1 = 3

a₅ = a₄ + a₃ = 3 + 2 = 5

a₆ = a₅ + a₄ = 5 + 3 = 8

Thus,

Exercise 8.2 Solutions

Q1. Find the 20^th and n^th terms of the G.P. 5/2, 5/4, 5/8, ………
Solution:
Given G.P. is 5/2, 5/4, 5/8, ………

Here, a = First term = 5/2

r = Common ratio = (5/4)/(5/2) = ½

Thus, the 20^th term and n^th term

Q2. Find the 12^th term of a G.P. whose 8^th 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,

a₈ = ar ^8–1 = ar⁷

ar⁷ = 192

a(2)⁷ = 192

a(2)⁷ = (2)⁶ (3)

Q3. The 5^th, 8^th and 11^th 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 = a r^5–1 = a r⁴ = p … (i)

a₈ = a r^8–1 = a r⁷ = q … (ii)

a₁₁ = a r^11–1 = a r¹⁰ = s … (iii)

Dividing equation (ii) by (i), we get

Q4. The 4^th term of a G.P. is the square of its second term, and the first term is –3. Determine its 7^th 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 = a r¹ = (–3) r

Then, from the question, we have

(–3) r³ = [(–3) r]²

⇒ –3r³ = 9 r²

⇒ r = –3

a₇ = a r ^7–1 = a r⁶ = (–3) (–3)⁶ = – (3)⁷ = –2187

Therefore, the seventh term of the G.P. is –2187.

Q5. Which term of the following sequences:

(a) 2, 2√2, 4,… is 128 ? (b) √3, 3, 3√3,… is 729 ?

(c) 1/3, 1/9, 1/27, … is 1/19683 ?
Solution: (a) The given sequence, 2, 2√2, 4,…

We have,

a = 2 and r = 2√2/2 = √2

Taking the n^th term of this sequence as 128, we have

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

(ii) Given the sequence, √3, 3, 3√3,…

We have,

a = √3 and r = 3/√3 = √3

Taking the n^th term of this sequence to be 729, we have

Therefore, the 12^th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, …

a = 1/3 and r = (1/9)/(1/3) = 1/3

Taking the n^th term of this sequence to be 1/19683, we have

Therefore, the 9^th term of the given sequence is 1/19683.

Q6. For what values of x, the numbers -2/7, x, -7/2 are in G.P?
Solution: The given numbers are -2/7, x, -7/2

Common ratio = x/(-2/7) = -7x/2

Also, common ratio = (-7/2)/x = -7/2x

Therefore, for x = ± 1, the given numbers will be in G.P.

Q7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …
Solution:
Given G.P., 0.15, 0.015, 0.00015, …

Here, a = 0.15 and r = 0.015/0.15 = 0.1 

Q8. Find the sum to n terms in the geometric progression √7, √21, 3√7, ….
Solution: The given G.P. is √7, √21, 3√7, ….

Here,

a = √7 and

Q9. Find the sum to n terms in the geometric progression 1, -a, a², -a³ …. (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,

Q10. Find the sum to n terms in the geometric progression x³, x⁵, x⁷, … (if x ≠ ±1 )
Solution:
Given G.P. is x³, x⁵, x⁷, …

Here, we have a = x³ and r = x⁵/x³ = x²

Q11. Evaluate:
Solution:

Q12. The sum of the first three terms of a G.P. is 39/10, and their product is 1. Find the common ratio and the terms.
Solution: Let a/r, a, ar be the first three terms of the G.P.

a/r + a + ar = 39/10 …… (1)

(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

1/r + 1 + r = 39/10

(1 + r + r²)/r = 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 = 2/5 or 5/2

Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.

Q13. 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. be required to obtain the sum 120.

We know that,

Here, a = 3 and r = 3

Equating the exponents, we get n = 4

Therefore, four terms of the given G.P. are required to obtain the sum 120.

Q14. 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

r³ = 8

r = 2

Now, using r = 2 in (1), we get

a (1 + 2 + 4) = 16

a (7) = 16

a = 16/7

Now, the sum of terms is given as

Q15. Given a G.P. with a = 729 and 7^th 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 = a r^n–1

a₇ = ar^7–1 = (729)r⁶

⇒ 64 = 729 r⁶

r⁶ = 64/729

r⁶ = (2/3)⁶

r = 2/3

And we know that

Q16. 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

And,

a₅ = 4 x a₃

ar⁴ = 4ar²

r² = 4

r = ± 2

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

Therefore, the required G.P is

-4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……

Q17. If the 4^th, 10^th and 16^th terms of a G.P. are x, y and z, respectively. 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 = a r³ = x … (1)

a₁₀ = a r⁹ = y … (2)

a₁₆ = a r¹⁵ = z … (3)

On dividing (2) by (1), we get

Q18. Find the sum to 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

Q19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.
Solution: The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½

= 64[4 + 2 + 1 + ½ + 1/2²]

Now, it’s seen that

4, 2, 1, ½, 1/2² is a G.P.

With the first term, a = 4

Common ratio, r =1/2

We know,

Therefore, the required sum = 64(31/4) = (16)(31) = 496

Q20. 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 be proved: The sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P.

Now, we have

Therefore, the above sequence forms a G.P., and the common ratio is rR.

Q21. 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 4^th 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 (1) and (2), we get

a(r² – 1) = 9 … (iii)

ar (1– r²) = 18 … (iv)

Now, dividing (4) by (3), we get

-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.

Q22. If the p^th, q^th and r^th 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

AR^p–1 = a

AR^q–1 = b

AR^r–1 = c

Then,

a^q–r b^r–p c^p–q

= A^q–r × R^{(p–1) (q–r)} × A^r–p × R^{(q–1) (r–p)} × A^p–q × R^{(r –1)(p–q)}

= Aq ^{– r + r – p + p – q} × R ^{(pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)}

= A⁰ × R⁰

= 1

Hence proved.

Q23. If the first and the n^th 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 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

= (a) (ar) (ar²) … (ar^n–1)

= (a × a ×…a) (r × r² × …r^n–1)

= aⁿ r ^{1 + 2 +…(n–1)}  … (2)

Here, 1, 2, …(n – 1) is an A.P.

And, the product of n terms P is given by,

Q24. Show that the ratio of the sum of the first n terms of a G.P. to the sum of terms from .
Solution: Let a be the first term and r be the common ratio of the G.P.

Since there are n terms from (n +1)^th to (2n)^th term,

Sum of terms from(n + 1)^th to (2n)^th term

a ^n +1 = ar ^{n + 1 – 1} = arⁿ

Thus, the required ratio =

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 ^.

Q25. If a, b, c and d are in G.P., show that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².
Solution: Given a, b, c, 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 +2a²bd + 2acbd + d²(a² + 2ac + c²)

= a²b² + 2a²c² + 2b²c² + d²a² + 2d²b² + d²c²  [Using (1) and (2)]

= a²b² + a²c² + a²c² + b²c² + b²c² + d²a² + d²b² + d²b² + d²c²

= a²b² + a²c² + a²d² + b² × b² + b²c² + b²d² + c²b² + c² × c² + 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

Q26. 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 1^st term as 3 and the 4^th 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 which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P are 9 and 27.

Q27. Find the value of n so that may be the geometric mean between a and b.
Solution: We know that,

The G. M. of a and b is given by √ab.

Then from the question, we have

By squaring both sides, we get

Q28. The sum of two numbers is 6 times their geometric mean; show that numbers are in the ratio.
Solution: Consider the two numbers to be a and b.

Then, G.M. = √ab.

From the question, we have

Q29. If A and G be A.M. and G.M., respectively, between two positive numbers, prove that the numbers are.
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.

Q30. 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 2^nd hour, 4^th hour and n^th hour?
Solution: Given 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 2^nd hour will be 120.

And, a₅ = ar⁴ = (30) (2)⁴ = 480

The number of bacteria at the end of 4^th hour will be 480.

a_n +1 = arⁿ = (30) 2ⁿ

Therefore, the number of bacteria at the end of n^th hour will be 30(2)ⁿ.

Q31. What will Rs 500 amount to in 10 years after its deposit in a bank which pays an annual interest rate of 10% compounded annually?
Solution:
Given,

The amount deposited in the bank is Rs 500.

At the end of first year, amount = Rs 500(1 + 1/10) = Rs 500 (1.1)

At the end of 2^nd year, amount = Rs 500 (1.1) (1.1)

At the end of 3^rd 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

Q32. If A.M. and G.M. of 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

We know that,

A quadratic equation can be formed as,

x² – x (Sum of roots) + (Product of roots) = 0

x² – x (a + b) + (ab) = 0

x² – 16x + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation is x² – 16x + 25 = 0