NCERT Solutions for Class 11 Science Math Chapter 9 - Sequences And Series

Answer:

Substituting n = 1, 2, 3, 4, and 5, we obtain

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

Answer:

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are.

Answer:

a_n = 2ⁿ

Substituting n = 1, 2, 3, 4, 5, we obtain

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

Answer:

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are.

Answer:

Substituting n = 1, 2, 3, 4, 5, we obtain

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

Answer:

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are

Answer:

Substituting n = 17, we obtain

Substituting n = 24, we obtain

Answer:

Here, $a_n=\frac{n^2}{2^n}$.
Substituting n = 7, we obtain

Answer:

Substituting n = 9, we obtain

Answer:

Substituting n = 20, we obtain

Answer:

Hence, the first five terms of the sequence are 3, 11, 35, 107, and 323.

The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Answer:

Hence, the first five terms of the sequence are

The corresponding series is

Answer:

Hence, the first five terms of the sequence are 2, 2, 1, 0, and –1.

The corresponding series is 2 + 2 + 1 + 0 + (–1) + …

Answer:

Answer:

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

This sequence forms an A.P.

Here, first term, a = 1

Common difference, d = 2

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

Answer:

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

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

Answer:

First term = 2

Let d be the common difference of the A.P.

Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, …

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

According to the given condition,

Thus, the 20^th term of the A.P. is –112.

Answer:

Let the sum of n terms of the given A.P. be –25.

It is known that, , where n = number of terms, a = first term, and d = common difference

Here, a = –6

Therefore, we obtain

Answer:

It is known that the general term of an A.P. is a_n = a + (n – 1)d

∴ According to the given information,

Subtracting (2) from (1), we obtain

Putting the value of d in (1), we obtain

Thus, the sum of first pq terms of the A.P. is .

Answer:

Let the sum of n terms of the given A.P. be 116.

Here, a = 25 and d = 22 – 25 = – 3

However, n cannot be equal to . Therefore, n = 8

∴ a₈ = Last term = a + (n – 1)d = 25 + (8 – 1) (– 3)

= 25 + (7) (– 3) = 25 – 21

= 4

Thus, the last term of the A.P. is 4.

Answer:

It is given that the k^th term of the A.P. is 5k + 1.

k^th term = a_k = a + (k – 1)d

∴ a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

Comparing the coefficient of k, we obtain d = 5

a – d = 1

⇒ a – 5 = 1

⇒ a = 6

Answer:

It is known that,

According to the given condition,

Comparing the coefficients of n² on both sides, we obtain

∴ d = 2 q

Thus, the common difference of the A.P. is 2q.

Answer:

Let a₁, a₂, and d₁, d₂ be the first terms and the common difference of the first and second arithmetic progression respectively.

According to the given condition,

Substituting n = 35 in (1), we obtain

From (2) and (3), we obtain

Thus, the ratio of 18^th term of both the A.P.s is 179: 321.

Answer:

Let a and d be the first term and the common difference of the A.P. respectively.

Here,

According to the given condition,

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

Answer:

Let a₁ and d be the first term and the common difference of the A.P. respectively.

According to the given information,

Subtracting (2) from (1), we obtain

$$\begin{gathered} \left(p - 1\right)d - \left(q - 1\right)d = \frac{2a}{p} - \frac{2b}{q} \\ \Rightarrow d\left(p - 1 - q + 1\right) = \frac{2aq - 2bp}{pq} \\ \Rightarrow d\left(p - q\right) = \frac{2aq - 2bp}{pq} \\ \Rightarrow d = \frac{2\left(aq - bp\right)}{pq\left(p - q\right)} .......\left(4\right) \end{gathered}$$

Subtracting (3) from (2), we obtain

Equating both the values of d obtained in (4) and (5), we obtain

$$\begin{gathered} \frac{aq - bp}{pq\left(p - q\right)} = \frac{br - qc}{qr\left(q - r\right)} \\ \Rightarrow \frac{aq - bp}{p\left(p - q\right)} = \frac{br - qc}{r\left(q - r\right)} \\ \Rightarrow r\left(q - r\right)\left(aq - bp\right) = p\left(p - q\right)\left(br - qc\right) \\ \Rightarrow r\left(aq - bp\right)\left(q - r\right) = p\left(br - qc\right)\left(p - q\right) \\ \Rightarrow \left(aqr - bpr\right)\left(q - r\right) = \left(bpr - cpq\right)\left(p - q\right) \end{gathered}$$

Dividing both sides by pqr, we obtain

Thus, the given result is proved.

Answer:

Let a and d be the first term and the common difference of the A.P. respectively.

According to the given condition,

Putting m = 2m – 1 and n = 2n – 1 in (1), we obtain

From (2) and (3), we obtain

Thus, the given result is proved.

Answer:

Let a and b be the first term and the common difference of the A.P. respectively.

a_m = a + (m – 1)d = 164 … (1)

Sum of n terms,

Here,

$$\begin{gathered} \frac{n}{2} \left[2a + nd - d\right] = 3n^2 + 5n \\ \Rightarrow na + \frac{d}{2}{n}^2 - \frac{d}{2}n = 3n^2 + 5n \\ \Rightarrow \frac{d}{2}{n}^2 + \left(a - \frac{d}{2}\right)n = 3n^2 + 5n \end{gathered}$$

Comparing the coefficient of n² on both sides, we obtain

Comparing the coefficient of n on both sides, we obtain

Therefore, from (1), we obtain

8 + (m – 1) 6 = 164

⇒ (m – 1) 6 = 164 – 8 = 156

⇒ m – 1 = 26

⇒ m = 27

Thus, the value of m is 27.

Answer:

Let A₁, A₂, A₃, A₄, and A₅ be five numbers between 8 and 26 such that

8, A₁, A₂, A₃, A₄, A₅, 26 is an A.P.

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

Therefore, 26 = 8 + (7 – 1) d

⇒ 6d = 26 – 8 = 18

⇒ d = 3

A₁ = a + d = 8 + 3 = 11

A₂ = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A₃ = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A₄ = a + 4d = 8 + 4 × 3 = 8 + 12 = 20

A₅ = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

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

Answer:

A.M. of a and b

According to the given condition,

Answer:

Let A₁, A₂, … A_m be m numbers such that 1, A₁, A₂, … A_m, 31 is an A.P.

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

∴ 31 = 1 + (m + 2 – 1) (d)

⇒ 30 = (m + 1) d

A₁ = a + d

A₂ = a + 2d

A₃ = a + 3d …

∴ A₇ = a + 7d

A_m–1 = a + (m – 1) d

According to the given condition,

Thus, the value of m is 14.

Answer:

The first installment of the loan is Rs 100.

The second installment of the loan is Rs 105 and so on.

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

The A.P. is 100, 105, 110, …

First term, a = 100

Common difference, d = 5

A₃₀ = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Thus, the amount to be paid in the 30^th installment is Rs 245.

Answer:

The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°.

It is known that the sum of all angles of a polygon with n sides is 180° (n – 2).

Answer:

The given G.P. is

Here, a = First term =

r = Common ratio =

Answer:

Common ratio, r = 2

Let a be the first term of the G.P.

∴ a₈ = ar ^8–1 = ar⁷

⇒ ar⁷ = 192

a(2)⁷ = 192

a(2)⁷ = (2)⁶ (3)

Answer:

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a₅ = a r^5–1 = a r⁴ = p … (1)

a₈ = a r^8–1 = a r⁷ = q … (2)

a₁₁ = a r^11–1 = a r¹⁰ = s … (3)

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

Dividing equation (3) by (2), we obtain

Equating the values of r³ obtained in (4) and (5), we obtain

Thus, the given result is proved.

Answer:

Let a be the first term and r be the common ratio of the G.P.

∴ a = –3

It is known that, a_n = ar^n–1

∴a₄ = ar³ = (–3) r³

a₂ = a r¹ = (–3) r

According to the given condition,

(–3) r³ = [­(–3) r]²

⇒ –3r³ = 9 r²

⇒ r = –3

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

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

Answer:

(a) The given sequence is

Here, a = 2 and r =

Let the n^th term of the given sequence be 128.

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

(b) The given sequence is

Here,

Let the n^th term of the given sequence be 729.

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

(c) The given sequence is

Here,

Let the n^th term of the given sequence be .

Thus, the 9^th term of the given sequence is .

Answer:

The given numbers are .

Common ratio =

Also, common ratio =

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

Answer:

The given G.P. is 0.15, 0.015, 0.00015, …

Here, a = 0.15 and

Answer:

The given G.P. is

Here,

Answer:

The given G.P. is

Here, first term = a₁ = 1

Common ratio = r = – a

Answer:

The given G.P. is

Here, a = x³ and r = x²

Answer:

The terms of this sequence 3, 3², 3³, … forms a G.P.

Substituting this value in equation (1), we obtain

Answer:

Let be the first three terms of the G.P.

From (2), we obtain

a³ = 1

⇒ a = 1 (Considering real roots only)

Substituting a = 1 in equation (1), we obtain

Thus, the three terms of G.P. are .

Answer:

The given G.P. is 3, 3², 3³, …

Let n terms of this G.P. be required to obtain the sum as 120.

Here, a = 3 and r = 3

∴ n = 4

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

Answer:

Let the G.P. be a, ar, ar², ar³, …

According to the given condition,

a + ar + ar² = 16 and ar³ + ar⁴ + ar⁵ = 128

⇒ a (1 + r + r²) = 16 … (1)

ar³(1 + r + r²) = 128 … (2)

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

Substituting r = 2 in (1), we obtain

a (1 + 2 + 4) = 16

⇒ a (7) = 16

Answer:

a = 729

a₇ = 64

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

It is known that, a_n = a r^n–1

a₇ = ar^7–1 = (729)r⁶

⇒ 64 = 729 r⁶

Also, it is known that,

Answer:

Let a be the first term and r be the common ratio of the G.P.

According to the given conditions,

a₅ = 4 × a₃

ar⁴ = 4ar²

⇒ r² = 4

∴ r = ± 2

From (1), we obtain

Thus, the required G.P. is

4, –8, 16, –32, …

Answer:

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)

Dividing (2) by (1), we obtain

Dividing (3) by (2), we obtain

∴

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

Answer:

The given sequence is 8, 88, 888, 8888…

This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as

S_n = 8 + 88 + 888 + 8888 + …………….. to n terms

Answer:

Required sum =

Here, 4, 2, 1, is a G.P.

First term, a = 4

Common ratio, r =

It is known that,

∴Required sum =

Answer:

It has to be proved that the sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P.

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

Answer:

Let a be the first term and r be the common ratio of the G.P.

a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³

By the given condition,

a₃ = a₁ + 9

⇒ ar² = a + 9 … (1)

a₂ = a₄ + 18

⇒ ar = ar³ + 18 … (2)

From (1) and (2), we obtain

a(r² ­­– 1) = 9 … (3)

ar (1– r²) = 18 … (4)

Dividing (4) by (3), we obtain

Substituting the value of r in (1), we obtain

4a = a + 9

⇒ 3a = 9

∴ a = 3

Thus, 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.

Answer:

Let A be the first term and R be the common ratio of the G.P.

According to the given information,

AR^p–1 = a

AR^q–1 = b

AR^r–1 = c

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

Thus, the given result is proved.

Answer:

The first term of the G.P is a and the last term is b.

Therefore, the G.P. is a, ar, ar², ar³, … ar^n–1, where r is the common ratio.

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.

∴1 + 2 + ……….+ (n – 1)

Thus, the given result is proved.

Answer:

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, 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 (2n)^th term is ^.

Answer:

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

Therefore,

bc = ad … (1)

b² = ac … (2)

c² = bd … (3)

It has to be proved that,

(a² + b² + c²) (b² + c² + d²) = (ab + bc – cd)²

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.

∴ L.H.S. = R.H.S.

∴

Answer:

Let G₁ and G₂ be two numbers between 3 and 81 such that the series, 3, G₁, G₂, 81, forms a G.P.

Let a be the first term and r be the common ratio of the G.P.

∴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

Thus, the required two numbers are 9 and 27.

Answer:

G. M. of a and b is .

By the given condition,

Squaring both sides, we obtain

Answer:

Let the two numbers be a and b.

G.M. =

According to the given condition,

Also,

Adding (1) and (2), we obtain

Substituting the value of a in (1), we obtain

Thus, the required ratio is.

Answer:

It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.

From (1) and (2), we obtain

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 obtain

(a – b)² = 4A² – 4G² = 4 (A²–G²)

(a – b)² = 4 (A + G) (A – G)

From (3) and (5), we obtain

Substituting the value of a in (3), we obtain

Thus, the two numbers are.

Answer:

It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P.

Here, a = 30 and r = 2

∴ a₃ = ar² = (30) (2)² = 120

Therefore, the number of bacteria at the end of 2^nd hour will be 120.

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

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

Answer:

The amount deposited in the bank is Rs 500.

At the end of first year, amount = = 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

∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)¹⁰

Answer:

Let the root of the quadratic equation be a and b.

According to the given condition,

The quadratic equation is given by,

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

x² – x (a + b) + (ab) = 0

x² – 16x + 25 = 0 [Using (1) and (2)]

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

Answer:

The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

n^th term, a_n = n ( n + 1)

Answer:

The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

n^th term, a_n = n ( n + 1) ( n + 2)

= (n² + n) (n + 2)

= n³ + 3n² + 2n

Answer:

The given series is 3 ×1² + 5 × 2² + 7 × 3² + …

n^th term, a_n = ( 2n + 1) n² = 2n³ + n²

Answer:

The given series is

n^th term, a_n =

Adding the above terms column wise, we obtain

Answer:

The given series is 5² + 6² + 7² + … + 20²

n^th term, a_n = ( n + 4)² = n² + 8n + 16

16^th term is (16 + 4)² = 20²2

Answer:

The given series is 3 × 8 + 6 × 11 + 9 × 14 + …

a_n = (n^th term of 3, 6, 9 …) × (n^th term of 8, 11, 14, …)

= (3n) (3n + 5)

= 9n² + 15n

Answer:

The given series is 1² + (1² + 2²) + (1² + 2² + 3² ) + …

a_n = (1² + 2² + 3² +…….+ n²)

$$\begin{gathered} = \frac{n\left(n+1\right)\left(2n+1\right)}{6} \\ =\frac{n\left(2n^2+3n+1\right)}{6} \\ =\frac{2n^3+3n^2+n}{6} \\ =\frac{1}{3}{n}^3+\frac{1}{2}{n}^2+\frac{1}{6}n \end{gathered}$$

Answer:

a_n = n (n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n

Answer:

a_n = n² + 2ⁿ

Consider

The above series 2, 2², 2³, … is a G.P. with both the first term and common ratio equal to 2.

Therefore, from (1) and (2), we obtain

Answer:

a_n = (2n – 1)² = 4n² – 4n + 1

Answer:

Let a and d be the first term and the common difference of the A.P. respectively.

It is known that the k^th term of an A. P. is given by

a_k = a + (k –1) d

∴ a_{m + n} = a + (m + n –1) d

a_{m – n} = a + (m – n –1) d

a_m = a + (m –1) d

∴ a_{m + n} + a_{m – n} = a + (m + n –1) d + a + (m – n –1) d

= 2a + (m + n –1 + m – n –1) d

= 2a + (2m – 2) d

= 2a + 2 (m – 1) d

=2 [a + (m – 1) d]

= 2a_m

Thus, the sum of (m + n)^th and (m – n)^th terms of an A.P. is equal to twice the m^th term.

Answer:

Let the three numbers in A.P. be a – d, a, and a + d.

According to the given information,

(a – d) + (a) + (a + d) = 24 … (1)

⇒ 3a = 24

∴ a = 8

(a – d) a (a + d) = 440 … (2)

⇒ (8 – d) (8) (8 + d) = 440

⇒ (8 – d) (8 + d) = 55

⇒ 64 – d² = 55

⇒ d² = 64 – 55 = 9

⇒ d = ± 3

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.

Thus, the three numbers are 5, 8, and 11.

Answer:

Let a and b be the first term and the common difference of the A.P. respectively.

Therefore,

From (1) and (2), we obtain

Hence, the given result is proved.

Answer:

The numbers lying between 200 and 400, which are divisible by 7, are

203, 210, 217, ­­­­­­­­… 399

∴First term, a = 203

Last term, l = 399

Common difference, d = 7

Let the number of terms of the A.P. be n.

∴ a_n = 399 = a + (n –1) d

⇒ 399 = 203 + (n –1) 7

⇒ 7 (n –1) = 196

⇒ n –1 = 28

⇒ n = 29

Thus, the required sum is 8729.

Answer:

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒100 = 2 + (n –1) 2

⇒ n = 50

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This forms an A.P. with both the first term and common difference equal to 5.

∴100 = 5 + (n –1) 5

⇒ 5n = 100

⇒ n = 20

The integers, which are divisible by both 2 and 5, are 10, 20, … 100.

This also forms an A.P. with both the first term and common difference equal to 10.

∴100 = 10 + (n –1) (10)

⇒ 100 = 10n

⇒ n = 10

∴Required sum = 2550 + 1050 – 550 = 3050

Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

Answer:

The two-digit numbers, which when divided by 4, yield 1 as remainder, are

13, 17, … 97.

This series forms an A.P. with first term 13 and common difference 4.

Let n be the number of terms of the A.P.

It is known that the n^th term of an A.P. is given by, a_n = a + (n –1) d

∴97 = 13 + (n –1) (4)

⇒ 4 (n –1) = 84

⇒ n – 1 = 21

⇒ n = 22

Sum of n terms of an A.P. is given by,

Thus, the required sum is 1210.

Answer:

It is given that,

f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)

f (1) = 3

Taking x = y = 1 in (1), we obtain

f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly,

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81

∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.

It is known that,

It is given that,

Thus, the value of n is 4.

Answer:

Let the sum of n terms of the G.P. be 315.

It is known that,

It is given that the first term a is 5 and common ratio r is 2.

∴Last term of the G.P = 6^th term = ar^{6 – 1} = (5)(2)⁵ = (5)(32) = 160

Thus, the last term of the G.P. is 160.

Answer:

Let a and r be the first term and the common ratio of the G.P. respectively.

∴ a = 1

a₃ = ar² = r²

a₅ = ar⁴ = r⁴

∴ r² + r⁴ = 90

⇒ r⁴ + r² – 90 = 0

Thus, the common ratio of the G.P. is ±3.

Answer:

Let the three numbers in G.P. be a, ar, and ar².

From the given condition, a + ar + ar² = 56

⇒ a (1 + r + r²) = 56

… (1)

a – 1, ar – 7, ar² – 21 forms an A.P.

∴(ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)

⇒ ar – a – 6 = ar² – ar – 14

⇒ar² – 2ar + a = 8

⇒ar² – ar – ar + a = 8

⇒a(r² + 1 – 2r) = 8

⇒ a (r – 1)² = 8 … (2)

⇒7(r² – 2r + 1) = 1 + r + r²

⇒7r² – 14 r + 7 – 1 – r – r² = 0

⇒ 6r² – 15r + 6 = 0

⇒ 6r² – 12r – 3r + 6 = 0

⇒ 6r (r – 2) – 3 (r – 2) = 0

⇒ (6r – 3) (r – 2) = 0

When r = 2, a = 8

When

Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.

When , the three numbers in G.P. are 32, 16, and 8.

Thus, in either case, the three required numbers are 8, 16, and 32.

Answer:

Let the G.P. be T₁, T₂, T₃, T₄, … T_2n.

Number of terms = 2n

According to the given condition,

T₁ + T₂ + T₃ + …+ T_2n = 5 [T₁ + T₃ + … +T_2n–1]

⇒ T₁ + T₂ + T₃ + … + T_2n – 5 [T₁ + T₃ + … + T_2n–1] = 0

⇒ T₂ + T₄ + … + T_2n = 4 [T₁ + T₃ + … + T_2n–1]

Let the G.P. be a, ar, ar², ar³, …

Thus, the common ratio of the G.P. is 4.

Answer:

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d]

+ [a + n – 1) d]

= 4a + (4n – 10) d

According to the given condition,

4a + 6d = 56

⇒ 4(11) + 6d = 56 [Since a = 11 (given)]

⇒ 6d = 12

⇒ d = 2

∴ 4a + (4n –10) d = 112

⇒ 4(11) + (4n – 10)2 = 112

⇒ (4n – 10)2 = 68

⇒ 4n – 10 = 34

⇒ 4n = 44

⇒ n = 11

Thus, the number of terms of the A.P. is 11.

Answer:

It is given that,

From (1) and (2), we obtain

Thus, a, b, c, and d are in G.P.

Answer:

Let the G.P. be a, ar, ar², ar³, … ar^{n – 1}…

According to the given information,

Hence, P² Rⁿ = Sⁿ

Answer:

Let t and d be the first term and the common difference of the A.P. respectively.

The n^th term of an A.P. is given by, a_n = t + (n – 1) d

Therefore,

a_p = t + (p – 1) d = a … (1)

a_q = t + (q – 1)d = b … (2)

a_r = t + (r – 1) d = c … (3)

Subtracting equation (2) from (1), we obtain

(p – 1 – q + 1) d = a – b

⇒ (p – q) d = a – b

Subtracting equation (3) from (2), we obtain

(q – 1 – r + 1) d = b – c

⇒ (q – r) d = b – c

Equating both the values of d obtained in (4) and (5), we obtain

Thus, the given result is proved.

Answer:

It is given that a are in A.P.

Thus, a, b, and c are in A.P.

Answer:

It is given that a, b, c,and d are in G.P.

∴b² = ac … (1)

c² = bd … (2)

ad = bc … (3)

It has to be proved that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P. i.e.,

(bⁿ + cⁿ)² = (aⁿ + bⁿ) (cⁿ + dⁿ)

Consider L.H.S.

(bⁿ + cⁿ)² = b²ⁿ + 2bⁿcⁿ + c²ⁿ

= (b²)ⁿ+ 2bⁿcⁿ + (c²) ⁿ

= (ac)ⁿ + 2bⁿcⁿ + (bd)ⁿ [Using (1) and (2)]

= aⁿ cⁿ + bⁿcⁿ+ bⁿ cⁿ + bⁿ dⁿ

= aⁿ cⁿ + bⁿcⁿ+ aⁿ dⁿ + bⁿ dⁿ [Using (3)]

= cⁿ (aⁿ + bⁿ) + dⁿ (aⁿ + bⁿ)

= (aⁿ + bⁿ) (cⁿ + dⁿ)

= R.H.S.

∴ (bⁿ + cⁿ)² = (aⁿ + bⁿ) (cⁿ + dⁿ)

Thus, (aⁿ + bⁿ), (bⁿ + cⁿ), and (cⁿ + dⁿ) are in G.P.

Answer:

It is given that a and b are the roots of x² – 3x + p = 0

∴ a + b = 3 and ab = p … (1)

Also, c and d are the roots of

∴c + d = 12 and cd = q … (2)

It is given that a, b, c, d are in G.P.

Let a = x, b = xr, c = xr², d = xr³

From (1) and (2), we obtain

x + xr = 3

⇒ x (1 + r) = 3

xr² + xr³ =12

⇒ xr² (1 + r) = 12

On dividing, we obtain

Case I:

When r = 2 and x =1,

ab = x²r = 2

cd = x²r⁵ = 32

Case II:

When r = –2, x = –3,

ab = x²r = –18

cd = x²r⁵ = – 288

Thus, in both the cases, we obtain (q + p): (q – p) = 17:15

Answer:

Let the two numbers be a and b.

A.M and G.M. =

According to the given condition,

Using this in the identity (a – b)² = (a + b)² – 4ab, we obtain

Adding (1) and (2), we obtain

Substituting the value of a in (1), we obtain

Answer:

It is given that a, b, c are in A.P.

∴ b – a = c – b … (1)

It is given that b, c, d, are in G.P.

∴ c² = bd … (2)

Also, are in A.P.

It has to be proved that a, c, e are in G.P. i.e., c² = ae

From (1), we obtain

From (2), we obtain

Substituting these values in (3), we obtain

Thus, a, c, and e are in G.P.

Answer:

(i) 5 + 55 + 555 + …

Let S_n = 5 + 55 + 555 + ….. to n terms

(ii) .6 +.66 +. 666 +…

Let S_n = 06. + 0.66 + 0.666 + … to n terms

Answer:

The given series is 2 × 4 + 4 × 6 + 6 × 8 + … n terms

∴ n^th term = a_n = 2n × (2n + 2) = 4n² + 4n

a₂₀ = 4 (20)² + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680

Thus, the 20^th term of the series is 1680.

Answer:

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + …+ a_n–1 + a_n

S = 3 + 7 + 13 + 21 + …. + a_{n – 2} + a_{n – 1} + a_n

On subtracting both the equations, we obtain

S – S = [3 + (7 + 13 + 21 + 31 + …+ a_n–1 + a_n)] – [(3 + 7 + 13 + 21 + 31 + …+ a_n–1) + a_n]

S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (a_n – a_n–1)] – a_n

0 = 3 + [4 + 6 + 8 + … (n –1) terms] – a_n

a_n = 3 + [4 + 6 + 8 + … (n –1) terms]

Answer:

From the given information,

Thus, from (1) and (2), we obtain

Answer:

The n^th term of the given series is

Answer:

n^th term of the numerator = n(n + 1)² = n³ + 2n² + n

n^th term of the denominator = n²(n + 1) = n³ + n²

From (1), (2), and (3), we obtain

Thus, the given result is proved.

Answer:

It is given that the farmer pays Rs 6000 in cash.

Therefore, unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

According to the given condition, the interest paid annually is

12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500

Thus, total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500

= 12% of (6000 + 5500 + 5000 + … + 500)

= 12% of (500 + 1000 + 1500 + … + 6000)

Now, the series 500, 1000, 1500 … 6000 is an A.P. with both the first term and common difference equal to 500.

Let the number of terms of the A.P. be n.

∴ 6000 = 500 + (n – 1) 500

⇒ 1 + (n – 1) = 12

⇒ n = 12

∴Sum of the A.P

Thus, total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)

= 12% of 39000 = Rs 4680

Thus, cost of tractor = (Rs 12000 + Rs 4680) = Rs 16680

Answer:

It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

∴Unpaid amount = Rs 22000 – Rs 4000 = Rs 18000

According to the given condition, the interest paid annually is

10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000

Thus, total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000

= 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of (1000 + 2000 + 3000 + … + 18000)

Here, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both equal to 1000.

Let the number of terms be n.

∴ 18000 = 1000 + (n – 1) (1000)

⇒ n = 18

∴ Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of Rs 171000 = Rs 17100

∴Cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

Answer:

The numbers of letters mailed forms a G.P.: 4, 4², … 4⁸

First term = 4

Common ratio = 4

Number of terms = 8

It is known that the sum of n terms of a G.P. is given by

It is given that the cost to mail one letter is 50 paisa.

∴Cost of mailing 87380 letters = Rs 43690

Thus, the amount spent when 8^th set of letter is mailed is Rs 43690.

Answer:

It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

∴ Interest in first year

∴Amount in 15^th year = Rs

= Rs 10000 + 14 × Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

Amount after 20 years =

= Rs 10000 + 20 × Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

Answer:

Cost of machine = Rs 15625

Machine depreciates by 20% every year.

Therefore, its value after every year is 80% of the original cost i.e., of the original cost.

∴ Value at the end of 5 years = = 5 × 1024 = 5120

Thus, the value of the machine at the end of 5 years is Rs 5120.

Answer:

Let x be the number of days in which 150 workers finish the work.

According to the given information,

150x = 150 + 146 + 142 + …. (x + 8) terms

The series 150 + 146 + 142 + …. (x + 8) terms is an A.P. with first term 150, common difference –4 and number of terms as (x + 8)

However, x cannot be negative.

∴x = 17

Therefore, originally, the number of days in which the work was completed is 17.

Thus, required number of days = (17 + 8) = 25