# Chapter – 9: Sequences and Series

# Exercise 9.1

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

**1. 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

a_{1} = 1(1 + 2) = 3

a_{2} = 2(2 + 2) = 8

a_{3} = 3(3 + 2) = 15

a_{4} = 4(4 + 2) = 24

a_{5} = 5(5 + 2) = 35

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

**2. a**_{n}** = n/n+1**

**Solution: –**

Given 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.

**3. a**

_{n}**= 2**

^{n}**Solution: –**

Given n^{th} term, *a** _{n}* = 2

^{n}On substituting* n* = 1, 2, 3, 4, 5, we get

a_{1} = 2^{1} = 2

a_{2} = 2^{2} = 4

a_{3} = 2^{3} = 8

a_{4} = 2^{4} = 16

a_{5} = 2^{5} = 32

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

4. *a*_{n}** = (2n – 3)/6**

**Solution: –**

Given 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..

**5. a**_{n}** = (-1)**^{n-1}** 5**^{n+1}

**Solution: –**

Given n^{th} term, a_{n} = (-1)^{n-1} 5^{n+1}

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

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

**6.**

**Solution: –**

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

**7. a**_{n}** = 4n – 3; a**_{17}**, a**_{24}

**Solution: –**

Given,

*n*^{th} term of the sequence is a_{n} = 4n – 3

On substituting *n* = 17, we get

a_{17} = 4(17) – 3 = 68 – 3 = 65

Next, on substituting *n* = 24, we get

a_{24} = 4(24) – 3 = 96 – 3 = 93

**8. a**_{n}** = n**^{2}**/2**^{n}** ; a**^{7}

**Solution: –**

Given,

*n*^{th} term of the sequence is a_{n} = n^{2}/2^{n}

Now, on substituting *n* = 7, we get

a_{7} = 7^{2}/2^{7} = 49/ 128

**9. a**_{n}** = (-1)**^{n-1}** n**^{3}**; a**_{9}

**Solution: –**

Given,

*n*^{th} term of the sequence is a_{n} = (-1)^{n-1} n^{3}

On substituting *n* = 9, we get

a_{9} = (-1)^{9-1} (9)^{3} = 1 x 729 = 729

**10.**

**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:**

**11. a**_{1}** = 3, a**_{n}** = 3a**_{n-1}** + 2 for all n > 1**

**Solution: –**

Given, a_{n} = 3a_{n-1} + 2 and a_{1} = 3

Then,

a_{2} = 3a_{1} + 2 = 3(3) + 2 = 11

a_{3} = 3a_{2} + 2 = 3(11) + 2 = 35

a_{4} = 3a_{3} + 2 = 3(35) + 2 = 107

a_{5} = 3a_{4} + 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 …….

**12. a**_{1}** = -1, a**_{n}** = a**_{n-1}**/n, n ≥ 2**

**Solution: –**

Given,

a_{n} = a_{n-1}/n and a_{1} = -1

Then,

a_{2} = a_{1}/2 = -1/2

a_{3} = a_{2}/3 = -1/6

a_{4} = a_{3}/4 = -1/24

a_{5} = a_{4}/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) + …….

**13. a**_{1}** = a**_{2 }**= 2, a**_{n}** = a**_{n-1}** – 1, n > 2**

**Solution: –**

Given,

a_{1} = a_{2}, a_{n} = a_{n-1} – 1

Then,

a_{3} = a_{2} – 1 = 2 – 1 = 1

a_{4} = a_{3} – 1 = 1 – 1 = 0

a_{5} = a_{4} – 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) + ……

**14. The Fibonacci sequence is defined by**

**1 = a**_{1}** = a**_{2}** 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 = a_{1} = a_{2}

a_{n} = a_{n – 1 }+ a_{n – 2}, n > 2

So,

a_{3} = a_{2} + a_{1} = 1 + 1 = 2

a_{4} = a_{3} + a_{2} = 2 + 1 = 3

a_{5} = a_{4} + a_{3} = 3 + 2 = 5

a_{6} = a_{5} + a_{4} = 5 + 3 = 8

Thus,

# Exercise 9.2

**1. Find the sum of odd integers from 1 to 2001.**

**Solution: –**

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, *a* = 1

Common difference, *d* = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

S_{n} = n/2 [2a + (n-1)d]

Therefore, the sum of odd numbers from 1 to 2001 is 1002001.

**2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.**

**Solution: –**

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where, the first term, *a* = 105

Common difference, *d* = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

S_{n} = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

**3. In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20**^{th}** term is –112.**

**Solution: –**

Given,

The first term (a) of an A.P = 2

Let’s assume *d* be the common difference of the A.P.

So, the A.P. will be 2, 2 + *d*, 2 + 2*d*, 2 + 3*d*, …

Then,

Sum of first five terms = 10 + 10*d*

Sum of next five terms = 10 + 35*d*

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a_{20} = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20^{th} term of the A.P. is –112.

**4. How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?**

**Solution: –**

Let’s consider the sum of *n* terms of the given A.P. as –25.

We known that,

S_{n} = n/2 [2a + (n-1)d]

where *n* = number of terms, *a* = first term, and *d* = common difference

So here, *a* = –6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

**5. In an A.P., if p**

^{th}**term is 1/q and**

*q*

^{th}**term is 1/p, prove that the sum of first**

*pq*terms is ½ (pq + 1) where p ≠ q.**Solution: –**

**6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term**

**Solution: –**

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, sum of certain number of terms of the A.P. is 116

The number of terms be n

So, we have

S_{n} = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n^{2}

3n^{2} – 53n + 232 = 0

3n^{2} – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, 8^{th} term is the last term of the A.P.

a_{8} = 25 + (8 – 1)(-3)

= 25 – 21

= 4

**7. Find the sum to n terms of the A.P., whose k**

^{th }**term is 5**

*k*+ 1.**Solution: –**

Given, the *k*^{th} term of the A.P. is 5*k* + 1.

*k*^{th} term = *a** _{k}* =

*a*+ (

*k*– 1)

*d*

And,

*a *+ (*k* – 1)*d* = 5*k* + 1

*a* + *kd* – *d* = 5*k* + 1

On comparing the coefficient of *k*, we get *d* = 5

*a *– *d *= 1

*a* – 5 = 1

⇒ *a* = 6

**8. If the sum of n terms of an A.P. is (pn + qn**

^{2}**), where**

*p*and*q*are constants, find the common difference.**Solution: –**

We know that,

S_{n} = n/2 [2a + (n-1)d]

From the question we have,

On comparing the coefficients of *n*^{2} on both sides, we get

d/2 = q

Hence, *d* = 2*q*

Therefore, the common difference of the A.P. is 2*q*.

**9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18**

^{th}**terms.**

**Solution: –**

Let *a*_{1}, *a*_{2}, and *d*_{1}, *d*_{2 }be the first terms and the common difference of the first and second arithmetic progression respectively.

Then, from the question we have

**10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.**

**Solution: –**

Let’s take *a* and *d* to be the first term and the common difference of the A.P. respectively.

Then, it given that

Therefore, the sum of (p + q) terms of the A.P. is 0.

**11. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.**

**Prove that **

**Solution: –**

Let *a*_{1} and *d* be the first term and the common difference of the A.P. respectively.

Then according to the question, we have

Now, subtracting (2) from (1), we get

**12. The ratio of the sums of m and n terms of an A.P. is m**

^{2}**:**

*n*

^{2}**. Show that the ratio of**

*m*

^{th }**and**

*n*

^{th}**term is (2**

*m*– 1): (2*n*– 1).**Solution: –**

Let’s consider that *a* and *b* to be the first term and the common difference of the A.P. respectively.

Then from the question, we have

Hence, the given result is proved.

**13. If the sum of n terms of an A.P. is 3n**

^{2}**+ 5n and its**

*m*

^{th }**term is 164, find the value of**

*m*.**Solution: –**

Let’s consider *a* and *b* to be the first term and the common difference of the A.P. respectively.

*a** _{m}* =

*a*+ (

*m*– 1)

*d*= 164 … (1)

We the sum of the terms is given by,

S_{n} = n/2 [2a + (n-1)d]

**14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.**

**Solution: –**

Let’s assume A_{1}, A_{2}, A_{3}, A_{4}, and A_{5} to be five numbers between 8 and 26 such that 8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26 are in an A.P.

Here we have,

*a *= 8, *b *= 26, *n* = 7

So,

26 = 8 + (7 – 1) *d*

6*d* = 26 – 8 = 18

*d *= 3

Now,

A_{1} = *a* + *d* = 8 + 3 = 11

A_{2} = *a* + 2*d* = 8 + 2 × 3 = 8 + 6 = 14

A_{3} = *a* + 3*d* = 8 + 3 × 3 = 8 + 9 = 17

A_{4} = *a* + 4*d *= 8 + 4 × 3 = 8 + 12 = 20

A_{5} = *a* + 5*d* = 8 + 5 × 3 = 8 + 15 = 23

Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

**15. If**

** is the A.M. between a and b, then find the value of n.**

**Solution: –**

The A.M between a and b is given by, (a + b)/2

Then according to the question,

Thus, the value of n is 1.

**16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7**

^{th}**and (**

*m*– 1)

^{th}**numbers is 5: 9. Find the value of**

*m*.**Solution: –**

Let’s consider a_{1}, a_{2}, … a* _{m}* be

*m*numbers such that 1, a

_{1}, a

_{2}, … a

*, 31 is an A.P.*

_{m}And here,

*a* = 1, *b* = 31, *n* = *m* + 2

So, 31 = 1 + (*m* + 2 – 1) (*d*)

30 = (*m* + 1) *d*

d = 30/ (m + 1) ……. (1)

Now,

a_{1} = *a* + *d*

a_{2} = *a* + 2*d*

a_{3} = *a* + 3*d* …

Hence, a_{7} = *a* + 7*d*

a_{m}_{–1} = *a* + (*m* – 1)* d*

According to the question, we have

Therefore, the value of m is 14.

**17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30**^{th}** instalment?**

**Solution: –**

Given,

The first instalment of the loan is Rs 100.

The second instalment of the loan is Rs 105 and so on as the instalment increases by Rs 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And the, A.P. is 100, 105, 110, …

Where, first term, *a* = 100

Common difference, *d* = 5

So, the 30^{th} term in this A.P. will be

A_{30} = *a* + (30 – 1)*d*

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30^{th} instalment will be Rs 245.

**18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.**

**Solution: –**

It’s understood from the question that, the angles of the polygon will form an A.P. with common difference *d* = 5° and first term *a* = 120°.

And, we know that the sum of all angles of a polygon with *n* sides is 180° (*n* – 2).

Thus, we can say

Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.

# Exercise 9.3

**1. 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

**2. 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,

*a*_{8} = *ar* ^{8–1} = *ar*^{7}

*ar*^{7} = 192

*a*(2)^{7} = 192

*a*(2)^{7} = (2)^{6} (3)

**3. The 5**^{th}**, 8**^{th}** and 11**^{th}** terms of a G.P. are p, q and s, respectively. Show that q**

^{2}**=**

*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

*a*_{5} = *a* *r*^{5–1 }= *a* r^{4} = *p* … (i)

*a*_{8 }= *a* *r*^{8–1 }= *a* *r*^{7} = *q* … (ii)

*a*_{11} = a *r*^{11–1 }= *a* *r*^{10} = *s *… (iii)

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

**4. The 4**^{th}** term of a G.P. is 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, *a*_{4 }= *ar*^{3} = (–3) *r*^{3}

*a*_{2} = *a r*^{1} = (–3) *r*

Then from the question, we have

(–3) *r*^{3} = [(–3) *r*]^{2}

⇒ –3*r*^{3} = 9 *r*^{2}

⇒ *r* = –3

*a*_{7} = *a* *r* ^{7–1 }= *a* *r*^{6} = (–3) (–3)^{6} = – (3)^{7} = –2187

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

**5. 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 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.

**6. 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.

**7. 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

**8. 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

**9. Find the sum to n terms in the geometric progression 1, -a, a**

^{2}**, -a**

^{3}**…. (if a ≠ -1)**

**Solution: –**

The given G.P. is 1, -a, a^{2}, -a^{3} ….

Here, the first term = *a*_{1} = 1

And the common ratio = *r* = – *a*

We know that,

**10. Find the sum to n terms in the geometric progression x**

^{3}**, x**

^{5}**, x**

^{7}**, … (if x ≠ ±1 )**

**Solution: –**

Given G.P. is x^{3}, x^{5}, x^{7}, …

Here, we have *a* = *x*^{3} and *r* = *x*^{5}/x^{3} = x^{2}

**11. Evaluate:**

** **

**Solution: –**

**12. The sum of 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

a^{3} = 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 + r^{2})/r = 39/10

10 + 10r + 10r^{2} = 39r

10r^{2} – 29r + 10 = 0

10r^{2} – 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.

**13. How many terms of G.P. 3, 3**^{2}**, 3**^{3}**, … are needed to give the sum 120?**

**Solution: –**

Given G.P. is 3, 3^{2}, 3^{3}, …

Let’s consider that *n* terms of this G.P. be required to obtain the sum of 120.

We know that,

Equating the exponents we get, *n* = 4

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

**14. The sum of 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*, *ar*^{2}, *ar*^{3}, …

Then according to the question, we have

*a *+ *ar* + *ar*^{2} = 16 and *ar*^{3 }+ *ar*^{4} + *ar*^{5 }= 128

*a* (1 + *r* + *r*^{2}) = 16 … (1) and,

*ar*^{3}(1 + *r* + *r*^{2}) = 128 … (2)

Dividing equation (2) by (1), we get

r^{3} = 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

**15. Given a G.P. with a = 729 and 7**

^{th}**term 64, determine S**

_{7}**.**

**Solution: –**

Given,

*a* = 729 and *a*_{7} = 64

Let *r* be the common ratio of the G.P.

Then we know that, *a** _{n}* =

*a r*

^{n}^{–1}

*a*_{7} = *ar*^{7–1} = (729)*r*^{6}

⇒ 64 = 729 *r*^{6}

*r*^{6} = 64/729

*r*^{6} = (2/3)^{6}

r = 2/3

And, we know that

**16. Find a G.P. for which 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, S_{2} = -4

Then, from the question we have

And,

a_{5} = 4 x a_{3}

ar^{4} = 4ar^{2}

r^{2} = 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, ……

**17. If the 4**^{th}**, 10**^{th}** and 16**^{th}** terms of a G.P. are x, y and z, respectively. Prove that x, y, 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,

*a*_{4} = *a* *r*^{3} = *x* … (1)

*a*_{10} = *a* *r*^{9} =* y* … (2)

*a*_{16}^{ }=* a r*^{15 }= *z* … (3)

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

**18. 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

**19. 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 + 1/2^{2}]

Now, it’s seen that

4, 2, 1, ½, 1/2^{2} is a G.P.

With first term, *a* = 4

Common ratio, *r* =1/2

We know,

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

**20. Show that the products of the corresponding terms of the sequences a, ar, ar**^{2}**, …ar**^{n-1}** and A, AR, AR**^{2}**, … AR**^{n-1}** form a G.P, and find the common ratio.**

**Solution: –**

To be proved: The sequence, *aA*, *arAR*, *ar*^{2}*AR*^{2}, …*ar*^{n}^{–1}*AR*^{n}^{–1}, forms a G.P.

Now, we have

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

**21. Find four numbers forming a geometric progression in which 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,

*a*_{1} = *a*, *a*_{2} = *ar*, *a*_{3} = *ar*^{2}, *a*_{4} = *ar*^{3}

From the question, we have

*a*_{3} = *a*_{1} + 9

*ar*^{2} = *a* + 9 … (i)

*a*_{2} = *a*_{4} + 18

*ar *= *ar*^{3} + 18 … (ii)

So, from (1) and (2), we get

*a*(*r*^{2} – 1) = 9 … (iii)

*ar *(1– *r*^{2}) = 18 … (iv)

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

-r = 2

r = -2

On substituting the value of *r* in (i), we get

4*a *= *a* + 9

3*a* = 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.

**22. 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)}

= *A*^{0} × *R*^{0}

= 1

Hence proved.

**23. If the first and the n**

^{th}**term of a G.P. are**

*a*ad*b*, respectively, and if*P*is the product of*n*terms, prove that*P*

^{2}**= (**

*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*, *ar*^{2}, *ar*^{3}, … *ar*^{n}^{–1}, where *r* is the common ratio.

Then,

*b* = *ar*^{n}^{–1} … (1)

*P* = Product of *n* terms

= (*a*) (*ar*) (*ar*^{2}) … (*ar*^{n}^{–1})

= (*a* × *a* ×…*a*) (*r* × *r*^{2} × …*r*^{n}^{–1})

= *a*^{n}^{ }*r* ^{1 + 2 +…(n–1)} … (2)

Here, 1, 2, …(*n* – 1) is an A.P.

**24. Show that the ratio of the sum of 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 (2*n*)^{th} term,

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

*a* ^{n }^{+1} = *ar* ^{n + 1} ^{– 1} = *ar*^{n}

Thus, required ratio =

Thus, the ratio of the sum of first *n* terms of a G.P. to the sum of terms from (*n* + 1)^{th} to (2*n*)^{th }term is^{ .}

**25. If a, b, c and d are in G.P. show that (a**

^{2}**+ b**

^{2}**+ c**

^{2}**)(b**

^{2}**+ c**

^{2}**+ d**

^{2}**) = (ab + bc + cd)**

^{2}**.**

**Solution: –**

Given, *a*, *b*, *c*, *d* are in G.P.

So, we have

*bc* = *ad* … (1)

*b*^{2} = *ac *… (2)

*c*^{2} = *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}

= *a*^{2}*b*^{2} + 2*abd* (*a* + *c*) + *d*^{2} (*a* + *c*)^{2}

= *a*^{2}*b*^{2} +2*a*^{2}*bd* + 2*acbd* + *d*^{2}(*a*^{2} + 2*ac* + *c*^{2})

= *a*^{2}*b*^{2} + 2*a*^{2}*c*^{2} + 2*b*^{2}*c*^{2} + *d*^{2}*a*^{2} + 2*d*^{2}*b*^{2} + *d*^{2}*c*^{2} [Using (1) and (2)]

= *a*^{2}*b*^{2} + *a*^{2}*c*^{2} + *a*^{2}*c*^{2} + *b*^{2}*c*^{2 }+ *b*^{2}*c*^{2} + *d*^{2}*a*^{2} + *d*^{2}*b*^{2} + *d*^{2}*b*^{2} + *d*^{2}*c*^{2}

= *a*^{2}*b*^{2} + *a*^{2}*c*^{2} + *a*^{2}*d*^{2 }+ *b*^{2 }× *b*^{2} + *b*^{2}*c*^{2} + *b*^{2}*d*^{2} + *c*^{2}*b*^{2} + *c*^{2 }× *c*^{2} + *c*^{2}*d*^{2}

[Using (2) and (3) and rearranging terms]

= *a*^{2}(*b*^{2} + *c*^{2} + *d*^{2}) + *b*^{2} (*b*^{2} + *c*^{2} + *d*^{2}) + *c*^{2} (*b*^{2}+ *c*^{2} + *d*^{2})

= (*a*^{2} + *b*^{2} + *c*^{2}) (*b*^{2} + *c*^{2} + *d*^{2})

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

(a^{2} + b^{2} + c^{2})(b^{2} + c^{2} + d^{2}) = (ab + bc + cd)^{2}

**26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.**

**Solution: –**

Let’s assume *G*_{1} and *G*_{2} to be two numbers between 3 and 81 such that the series 3, *G*_{1}, *G*_{2}, 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}

*r*^{3} = 27

∴ *r* = 3 (Taking real roots only)

For *r* = 3,

*G*_{1} = *ar* = (3) (3) = 9

*G*_{2} = *ar*^{2} = (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.

**27. 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

**28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio.**

**Solution: –**

Consider the two numbers be *a* and *b*.

Then, G.M. = √ab.

From the question, we have

**29. 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*.

**30. 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 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, *a*_{3} = *ar*^{2} = (30) (2)^{2} = 120

Thus, the number of bacteria at the end of 2^{nd} hour will be 120.

And, *a*_{5} = *ar*^{4} = (30) (2)^{4} = 480

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

*a*_{n}_{ +1 }= *ar** ^{n}* = (30) 2

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

**31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays 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}

**32. 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,

*x*^{2 }– *x* (Sum of roots) + (Product of roots) = 0

*x*^{2} – *x* (*a* + *b*) + (*ab*) = 0

*x*^{2} – 16*x* + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation is *x*^{2} – 16*x* + 25 = 0

# Exercise 9.4

**Find the sum to n terms of each of the series in Exercises 1 to 7.**

**1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …**

**Solution: –**

Given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

It’s seen that,

*n*^{th} term, *a** _{n}* =

*n*(

*n*+ 1)

Then, the sum of n terms of the series can be expressed as

**2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …**

**Solution: –**

Given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

It’s seen that,

*n*^{th} term, *a** _{n}* =

*n*(

*n*+ 1) (

*n*+ 2)

= (*n*^{2} + *n*) (*n* + 2)

=* n*^{3 }+ 3*n*^{2 }+ 2*n*

Then, the sum of n terms of the series can be expressed as

**3. 3 × 1**^{2}** + 5 × 2**^{2}** + 7 × 3**^{2}** + …**

**Solution: –**

Given series is 3 ×1^{2} + 5 × 2^{2} + 7 × 3^{2} + …

It’s seen that,

*n*^{th} term, *a** _{n}* = (2

*n*+ 1)

*n*

^{2}= 2

*n*

^{3}+

*n*

^{2}

Then, the sum of n terms of the series can be expressed as

**4. Find the sum to n terms of the series **

**Solution: –**

**5. Find the sum to n terms of the series 5**

^{2}**+ 6**

^{2}**+ 7**

^{2}**+ … + 20**

^{2}**Solution: –**

Given series is 5^{2} + 6^{2} + 7^{2} + … + 20^{2}

It’s seen that,

*n*^{th} term, *a** _{n}* = (

*n*+ 4)

^{2}=

*n*

^{2}+ 8

*n*+ 16

Then, the sum of n terms of the series can be expressed as

**6. Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…**

**Solution: –**

Given series is 3 × 8 + 6 × 11 + 9 × 14 + …

It’s found out that,

*a** _{n }*= (

*n*

^{th}term of 3, 6, 9 …) × (

*n*

^{th}term of 8, 11, 14, …)

= (3*n*) (3*n* + 5)

= 9*n*^{2} + 15*n*

Then, the sum of n terms of the series can be expressed as

**7. Find the sum to n terms of the series 1**

^{2}**+ (1**

^{2}**+ 2**

^{2}**) + (1**

^{2}**+ 2**

^{2}**+ 3**

^{2}**) + …**

**Solution: –**

Given series is 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2 }+ 3^{2} ) + …

Finding the n^{th} term, we have

*a** _{n}* = (1

^{2}+ 2

^{2}+ 3

^{2}+…….+

*n*

^{2})

Now, the sum of n terms of the series can be expressed as

**8. Find the sum to n terms of the series whose n**

^{th}**term is given by**

*n*(*n*+ 1) (*n*+ 4).**Solution: –**

Given,

*a** _{n}* =

*n*(

*n*+ 1) (

*n*+ 4) =

*n*(

*n*

^{2 }+ 5

*n*+ 4) =

*n*

^{3}+ 5

*n*

^{2}+ 4

*n*

Now, the sum of n terms of the series can be expressed as

**9. Find the sum to n terms of the series whose n**

^{th}**terms is given by**

*n*

^{2}**+ 2**

^{n}**Solution: –**

Given,

n^{th} term of the series as:

*a*_{n}* = n*^{2} + 2^{n}

Then, the sum of n terms of the series can be expressed as

**10. Find the sum to n terms of the series whose n**

^{th}**terms is given by (2**

*n*– 1)

^{2}**Solution: –**

Given,

n^{th} term of the series as:

*a** _{n}* = (2

*n*– 1)

^{2}= 4

*n*

^{2}– 4

*n*+ 1

Then, the sum of n terms of the series can be expressed as

##### Tags In

### Related Posts

### Leave a Reply Cancel reply

**Error:** Contact form not found.