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(i) \(\displaystyle\sum_{n=1}^{11} (2 + 3^n)\)

= (2 + 3¹) + (2 + 3²) + (2 + 3³) + … + (2 + 3¹¹)

= 2×11 + 3¹ + 3² + 3³ + … + 3¹¹

= 22 + 3(3¹¹ – 1)/(3 – 1) [by using the formula, a(1 – rⁿ )/(1 – r)]

= 22 + 3(3¹¹ – 1)/2

= [44 + 3(177147 – 1)]/2

= [44 + 3(177146)]/2

= 265741

(ii) \(\displaystyle\sum_{k=1}^{n} (2^k + 3^{k - 1})\)

= (2 + 3⁰) + (2² + 3) + (2³ + 3²) + … + (2ⁿ + 3^n-1)

= (2 + 2² + 2³ + … + 2ⁿ) + (3⁰ + 3¹ + 3² + …. + 3^n-1)

Firstly let us consider,

(2 + 2² + 2³ + … + 2ⁿ)

Where, a = 2, r = 2²/2 = 4/2 = 2, n = n

By using the formula,

Sum of GP for n terms = a(rⁿ – 1 )/(r – 1)

= 2 (2ⁿ – 1)/(2 – 1)

= 2 (2ⁿ – 1)

Now, let us consider

(3⁰ + 3¹ + 3² + …. + 3ⁿ)

Where, a = 3⁰ = 1, r = 3/1 = 3, n = n

By using the formula,

Sum of GP for n terms = a(rⁿ – 1 )/(r – 1)

= 1 (3ⁿ – 1)/ (3 – 1)

= (3ⁿ – 1)/2

So,

\(\displaystyle\sum_{k=1}^{n} (2^k + 3^{k - 1})\)

= (2 + 2² + 2³ + … + 2ⁿ) + (3⁰ + 3¹ + 3² + …. + 3ⁿ)

= 2 (2ⁿ – 1) + (3ⁿ – 1)/2

= 1/2 [2ⁿ⁺² + 3ⁿ – 4 – 1]

= 1/2 [2ⁿ⁺² + 3ⁿ – 5]

(iii) \(\displaystyle\sum_{n=2}^{10} 4^n\)

= 4² + 4³ + 4⁴ + … + 4¹⁰

Where, a = 4² = 16, r = 4³/4² = 4, n = 9

By using the formula,

Sum of GP for n terms = a(rⁿ – 1 )/(r – 1)

= 16 (4⁹ – 1)/(4 – 1)

= 16 (4⁹ – 1)/3

= 16/3 [4⁹ – 1]

Let the five numbers in A.P. be a – 2d, a – d, a, a + d, a + 2d. 

Since 1st, 3rd and 4th terms are in G.P. 

(t₃)² = t₁ . t₄ 

⇒ a² = (a – 2d) (a + d) 

⇒ a² = a² – ad – 2d² 

⇒ – ad = 2d² ⇒ a = – 2d            (∵ d ≠ 0) 

⇒ a + 2d = 0 ⇒ t₅ = 0

∴ 5th term is always zero.

Let T indicate a term of the progression.T₁, T₂, T₃, ..., T_n, ...T_2nT₁ = 1 T₂ = a T₃ = ca T₄ = c.a² T₅ = c².a² T_k if k is even = \(\frac{k}{a^2}.\frac{k}{C^2}-1\)

T_2n = \(\frac{2n}{a^2}\). \(\frac{2n}{C^2}-1\)

T_2n = aⁿ. C^n-1

S_2n = 1 + a + ca + c.a² + c².a² + c².a³ .....aⁿ. c^{n - 1}= 1 + [ a + c.a² + c².a³.... + aⁿ.c^{n - 1}]

+ [ca + c².a² + c³.a³..... + c^{n - 1}.a^{n - 1}]The sum of a G.P. = \(\frac{a(r^{n-1})}{r-1}\)

For a + c.a² + c².a³.... + aⁿ.c^{n - 1} 

a = a, r = ca, n = n 

⇒ \(\frac{a(ca^n-1)}{ca-1}\)

For [ ca + c².a² + c³.a³..... + c^{n - 1}.a^{n - 1}] 

a = ca, r = ca, n = n

⇒ \(\frac{ca(ca^n-1)}{ca-1}\)

∴ The required result = 1 + \(\frac{a(ca^n-1)}{ca-1}\) + \(\frac{ca(ca^n-1)}{ca-1}\)

⇒ \(\cfrac{a(ca^n-1)+ca(ca^{n-1}) + ca -1}{ca-1}\)

Given GP is 0.4, 0.8, 1.6…. 

The given GP is of the form, a, ar, ar² , ar³…. 

Where r is the common ratio. 

First term in the given GP, a₁ = a = 0.4 

Second term in GP, a₂ = 0.8

Now, the common ratio, \(r = \frac{a_2}{a_1}\)

r = \(\frac{0.8}{0.4}\) = 2

Now, n^th term of GP is, an = ar^{n – 1} 

So, the 7^th term in the GP, 

a₇ = ar⁶ 

= 0.4 x 26 = 25.6 

n^th term in the GP, an = ar^{n – 1} = (0.4)(2)^{n – 1} 

= (0.2)2ⁿ 

Hence, 7^th term = 25.6 and n^th term = (0.2)2ⁿ

Given: 5^th term of a GP is 2. 

To find: the product of its first nine terms. 

First term is denoted by a, the common ratio is denote by r. 

∴ ar⁴ = 2 

We have to find the value of: a × ar¹ × ar² × ar³ × … × ar⁸ 

= a⁹r^{1 + 2 + 3 + 4 + … + 8} 

= a⁹r³⁶ 

= (ar⁴)⁹ 

= (2)⁹ 

= 512 .

(c) 12

Let a and r be the first term and common ratio respectively of the given G.P. Then, t₂ = ar, t₄ = ar³, t₅ = ar⁴, t₆ = ar⁵ 

Given, t₂ + t₅ = ar + ar⁴ = 216                 ...(i) 

and \(\frac{t_4}{t_6}\) = \(\frac{ar^3}{ar^5}\) = \(\frac{1}{4}\) ⇒ \(\frac{1}{r^2}\) = \(\frac{1}{4}\) ⇒ \(\frac{1}{r}\) = \(\frac{1}{2}\)       ...(ii)

Now putting r = 2, in (i) we get 

2a + 16a = 216 ⇒ 18a = 216 ⇒ a = 12.

Given: Second, third and sixth terms of an A.P. are consecutive terms of a G.P.

Let the first term of AP be a and the common difference be d.

⇒ A_n = a+(n-1)d

⇒ A₂ = a+d

⇒ A₃ = a+2d

⇒ A₆ = a+5d

If a,b,c are consecutive terms of GP then we can write b² = a.c

∴ We can write (a+2d)² = (a+d).(a+5d)

⇒ a²+4d²+4ad = a²+6ad+5d²

⇒ d²+2ad = 0

⇒ d(d+2a) = 0

∴ d = 0 or d = - 2a

When d = 0 then the GP becomes a, a, a.

∴ The common ration becomes 1.

When d = - 2a then the GP becomes – a, - 3a,- 9a

∴ The common ratio becomes 3.

Let the three numbers be \(\frac{a}{r}, a, ar.\)

⇒ \(\frac{a}{r}\) + a + ar = 14

⇒ a + ar + ar² = 14r…(1)

First two terms are increased by 1, and third decreased by 1

\(\therefore\) \(\frac{a}{r}\) + 1, a + 1, ar -1

The above sequence is in AP.

We know in AP.

2b = a + c

\(\therefore\) 2(a + 1) = ar - 1 + \(\frac{a}{r} + 1\)

⇒ 2a + 2 = \(\frac{ar^2 + a}{r}\)

⇒ 2ar + 2r = ar² + a 

⇒ ar² – 2ar + a = 2r …(2)

Dividing 1 by 2 we get

⇒ \(\frac{a + ar + ar^2}{ar^2 - 2ar + a}\) = \(\frac{14r}{2r}\)

⇒ \(\frac{1+r+r^2}{r^2 - 2r + 1} = 7\)

⇒ 1 + r + r² = 7r² – 14r + 7 

⇒ 6r² – 15r – 6 = 0 

⇒ 6r² – 12r – 3r – 6 = 0 

⇒ 6r(r – 2) – 3(r – 2) = 0 

⇒ (6r – 3) (r – 2) = 0 

⇒ r = 2 or r = 1/2. 

Substituting r = 2 in 2 we get 

⇒ a(2)² – 2a(2) + a = 2(2) 

⇒ 4a – 4a + a = 4 

⇒ a = 4

Substituting r = 1/2 in 2 we get 

⇒ a(1/2)² – 2a(1/2) + a = 2(1/2) 

⇒ a = 4

∴ substituting a and r we get the numbers as 2,4,8.

To find: 

The amount after three days Given: 

(i) Principal – 80000 

(ii) Time – 3 days 

(iii) Rate – 15% per annum 

Deduction = P × R × T

= 80000 \(\times \frac{15}{100} \times \frac{3}{365}\)

= 98.63 

The final amount after deduction = 80000 – 98.63

= 79901.37 

The value of the machine after 3 days is Rs. 79901.37

The nth term from the end is given by:

a_n = l (1/r)^n-1 where, l is the last term, r is the common ratio, n is the nth term

Given: last term, l = 162

r = t₂/t₁ = (2/9) / (2/27)

= 2/9 × 27/2

= 3

n = 4

So, a_n = l (1/r)^n-1

a₄ = 162 (1/3)^4-1

= 162 (1/3)³

= 162 × 1/27

= 6

∴ 4^th term from last is 6.

Let the three numbers be a/r, a, ar

So, according to the question

a/r + a + ar = 65 … equation (1)

a/r × a × ar = 3375 … equation (2)
From equation (2) we get,

a³ = 3375

a = 15.

From equation (1) we get,

(a + ar + ar²)/r = 65
a + ar + ar² = 65r … equation (3)

Substituting a = 15 in equation (3) we get

15 + 15r + 15r² = 65r

15r² – 50r + 15 = 0… equation (4)

Dividing equation (4) by 5 we get

3r² – 10r + 3 = 0

3r² – 9r – r + 3 = 0

3r(r – 3) – 1(r – 3) = 0

r = 3 or r = 1/3

Now, the equation will be

15/3, 15, 15×3 or

15/(1/3), 15, 15×1/3

So the terms are 5, 15, 45 or 45, 15, 5
∴ The three numbers are 5, 15, 45.

t₄ = ar³ = 54             ...(i) 

t₇ = ar⁶ = 1458          ...(ii)

∴ Dividing (ii) by (i), we get

\(\frac{t_7}{t_4}\) = r³ = \(\frac{1458}{54}\) = 27

⇒ r = 3

∴ From (i), a.(3)³ = 54 

⇒ 27a = 54 

⇒ a = 2

∴ The G.P. is 2, 2 x 3, 2 x 3², ..... , i.e. 2, 6, 18, 54, .... .

The nth term from the end is given by:

a_n = l (1/r)^n-1 where, l is the last term, r is the common ratio, n is the nth term

Given: last term, l = 1/4374

r = t₂/t₁ = (1/6) / (1/2)

= 1/6 × 2/1

= 1/3

n = 4

So, a_n = l (1/r)^n-1

a₄ = 1/4374 (1/(1/3))^4-1

= 1/4374 (3/1)³

= 1/4374 × 3³

= 1/4374 × 27

= 1/162

∴ 4^th term from last is 1/162.

(c) \(\frac{\sqrt5-1}{2}\)

Let the G.P. be a, ar, ar², ...... 

As all the terms of the given G.P. are positive, a > 0, r > 0. 

Given, a = ar + ar² 

⇒ ar² + ar – a = 0 

⇒ r² + r – 1 = 0.

∴ r = \(\frac{-1\pm\sqrt{1-4}}{2}\) = \(\frac{-1\pm\sqrt{5}}{2}\)

⇒ r = \(\frac{\sqrt{5-1}}{2}\).            \(\big(\because\frac{-1-\sqrt5}{2}\,\text{is a negative quantity and }\,r>0\big)\)

(b) – 12

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

Then a + ar = 12                  ...(i) 

ar² + ar³ = 48                     ....(ii)

⇒ \(\frac{ar^2(1+r)}{a(1+r)}\) = \(\frac{48}{12}\)            (Dividing (ii) by (i)) 

⇒ r² = 4 ⇒ r ± 2 ⇒ r = – 2 

as the terms of the G.P. are alternately positive and negative. 

Now a (1 + r) = 12 ⇒ a (1 – 2) = 12 ⇒ a = – 12.

By using the formula,

T_n = ar^n-1

a = 18

r = t₂/t₁ = (-12/18)

= -2/3

T_n = 512/729

n = ?

T_n = ar^n-1

512/729 = 18 (-2/3)^n-1

2⁹/(729 × 18) = (-2/3)^n-1

2⁹/36 × 1/2×3² = (-2/3)^n-1

(2/3)⁸ = (-1)^n-1 (2/3)^n-1

8 = n – 1

n = 8 + 1

= 9

∴ 9^th term of the Progression is 512/729

T_n = ar^n-1

a = 18, r =  \(\frac{-12}{18}\) = \(\frac{-2}{3}\), T_n = \(\frac{512}{729}, n = ?\)

\(\therefore\)\(\frac{512}{729}\) = 18 x (\(\frac{-2}{3}\))^n-1

^{\(\Rightarrow\)} \(\frac{512}{729}\) = 18 x  (\(\frac{-2}{3}\))ⁿ x \(\frac{-3}{2}\)

^{\(\Rightarrow\)}\(\frac{512}{19683}\) = \((\frac{2}{3})^3\)

^{\(\Rightarrow\) n = 9}

(b) 2059

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

Then, T₁ = a = 729, T₇ = ar⁶ = 64

∴ \(\frac{ar^6}{a}\) = \(\frac{64}{729}\) ⇒ r⁶ = \(\frac{64}{729}\) = \(\frac{2^6}{3^6}\) ⇒ r = \(\frac{2}{3}.\)

∴ S₇ = \(\frac{a(1-r^n)}{1-r}\) = \(\frac{729\bigg(1-\big(\frac{2}{3}\big)^7\bigg)}{1-\frac{2}{3}}\)

= \(\frac{729\bigg(1-\frac{128}{2187}\bigg)}{\frac{1}{3}}\) = 2187 \(\bigg(\frac{2187-128}{2187}\bigg)\) = 2059.

Given:

Sum of GP = 381

Where, a = 3, r = 6/3 = 2, n = ?

By using the formula,

Sum of GP for n terms = a(rⁿ – 1 )/(r – 1)

381 = 3 (2ⁿ – 1)/ (2 - 1)

381 = 3 (2ⁿ – 1)

381/3 = 2ⁿ – 1

127 = 2ⁿ – 1

127 + 1 = 2ⁿ

128 = 2ⁿ

2⁷ = 2ⁿ

n = 7

∴ value of n is 7

Given:

Sum of GP = 39 + 13√3

Where, a =√3, r = 3/√3 = √3, n = ?

By using the formula,

Sum of GP for n terms = a(rⁿ – 1 )/(r – 1)

39 + 13√3 = √3 (√3ⁿ – 1)/ (√3 – 1)

(39 + 13√3) (√3 – 1) = √3 (√3ⁿ – 1)

Let us simplify we get,

39√3 – 39 + 13(3) – 13√3 = √3 (√3ⁿ – 1)

39√3 – 39 + 39 – 13√3 = √3 (√3ⁿ – 1)

39√3 – 39 + 39 – 13√3 = √3ⁿ⁺¹ – √3

26√3 + √3 = √3ⁿ⁺¹

27√3 = √3ⁿ⁺¹

√3⁶ √3 = √3ⁿ⁺¹

6+1 = n + 1

7 = n + 1

7 – 1 = n

6 = n

∴ 6 terms are required to make a sum of 39 + 13√3

T_n = ar^n-1 

a =a, r =?, T_n = 27 n = 4 

a =a, r =?, T_n = 729 n = 7 

∴ 27 = a.r^4-1 

⇒ 27 =a.r³…(1) 

∴ 729 = a.r^7-1 

⇒ 729 = a.r⁶…(2) 

Divide (2) by (1) we get 

⇒\(\frac{729}{27}\)  = \(\frac{ar^6}{ar^3}\)

⇒ r³ = 27 

⇒ r = 3 

Substituting r in 1 we get 

a = 1 

∴ GP = 1,3,9…

Let G₁, G₂, G₃ be the required means.

Then 16, G₁, G₂, G₃, 256 form a G.P. 

Let r be the common ratio. 

⇒ 256 = 5th term = ar⁴ = 16 × r⁴ 

⇒ 16r⁴ = 256 ⇒ r⁴ = 16 ⇒ r = 2

∴ G₁ = ar = 16 x 2 = 32

G² = ar² = 16 x 4 = 64 

G³ = ar³ = 16 x 8 = 128 

Hence 32, 64 and 128 are the required G.M’s between 16 and 256.

Let the five terms be a₁, a₂, a₃, a₄, a₅.

A = 27, B = 1/81

Now, these 5 terms are between A and B.

So the GP is: A, a₁, a₂, a₃, a₄, a₅, B.

So we now have 7 terms in GP with the first term being 16 and seventh being 1/4.

We know that, T_n = ar^n–1

Here, T_n = 1/4, a = 16 and

1/4 = 16r^7-1

1/(4 × 16) = r⁶

r = 1/2

a₁ = Ar = 16 × 1/2 = 8  

a₂ = Ar² = 16 × 1/4 = 4   

a₃ = Ar³ = 16 × 1/8 = 2   

a₄ = Ar⁴ = 16 × 1/16 = 1   

a₅ = Ar⁵ = 16 × 1/32 = 1/2   

∴ The five GM between 16 and 1/4 are 8, 4, 2, 1, 1/2

Let the GM be y,z 

∴ 1,y,z,64 

y² = 1.z ; z² =64y 

y⁴ = z² 

∴ y⁴ =64y 

⇒ y = 4 

∴ Z =16

∴ The two GM are 4,16.

Let the GM be y,z

∴ 1,y,z,64

y² = 1.z ; z² = 64y

y⁴ = z²

∴ y⁴ = 64y

⇒ y = 4

∴ Z = 16

∴ The two GM are 4,16.

Given GP is √3, 3, 3√3…. 

The given GP is of the form, a, ar, ar² , ar³…. 

Where r is the common ratio. 

First term in the given GP, 

a₁ = a = √3 

Second term in GP, a₂ = 3 

Now, the common ratio, \(r = \frac{a_2}{a_1}\)

r = \(\frac{3}{√3}\) = √3

Let us consider 729 as the n^th term of the GP. Now, nth term of GP is, aⁿ = ar^{n – 1} 

729 = √3 (√3)^{n – 1} 

√3ⁿ = √3¹² 

n = 12 

So, 729 is the 12^th term in GP.

t₈ = 768

= ar⁷

r = 2

t₁₀ = ar⁹ 

= ar⁷ × r × r

= 768 × 2 × 2

= 3072

The nth term of a GP is an = ar^n-1 

It’s given in the question that 5th term of the GP is 80 and 8th term of GP is 640. 

So, a₅ = ar⁴ = 80 → (1)

a₈ = ar⁷ = 640 → (2)

\(\cfrac{(2)}{(1)}_\longrightarrow \cfrac{ar^7}{ar^4}\) = r³ = \(\cfrac{640}{80}\)= 8

Common ratio, r = 2, 

ar⁴ = 80 

16a = 80 a = 5 

The required GP is of the form a, ar, ar² , ar³ , ar⁴…. 

First term of GP, a = 5 

Second term of GP, ar = 5 x 2 =10 

Third term of GP, ar² = 5 x 2² = 20

Fourth term of GP, ar³ = 5 x 2³ = 40 

Fifth term of GP, ar⁴ = 5 x 2⁴ = 80 

And so on... 

The required GP is 5, 10, 20, 40, 80…

⇒ Let the first term be a and the common ratio be r.

∴ According to the question,

a_m+n = p.

a_m-n = q.

a_n = ar^n-1

a_m+n = a.r^m+n-1

a_m-n = a.r^m-n-1

∴ a.r^m+n-1 = p.

a.r^m-n-1 = n.

Multiplying above two equations we get

a²r^{(m+n-1+(m-n-1)} = a²r^(2m-2)

a²r^(2m-2) = p.q

(ar)^2(m-1) = p.q

∴ ar^m-1 =√p.q

⇒ M^th term is given by a.r^m-1

∴ ar^m-1 =√p.q

Let the first three terms of G.P. be \(\frac{a}{r}\), a, ar

It is  given that \(\frac{a}{r} \times a \times ar = 1\)

\(\Rightarrow\) a³ = 1

⇒ a = 1

And

\(\frac{a}{r} + a + ar = \frac{39}{10}\)

\(\Rightarrow\) a\(\left(\frac{1}{r} + 1 + r\right)\) = \(\frac{39}{10}\)

\(\Rightarrow\)\(\left(\frac{1}{r} + 1 + r\right)\) = \(\frac{39}{10}\) .....(a=1)

\(\Rightarrow\) \(\left(\frac{1}{r} + r\right)\) = \(\frac{39}{10}\) - 1 = \(\frac{29}{10}\)

⇒ 10(1 + r² ) = 29r 

⇒ 10r² - 29r + 10 = 0 

⇒ 10r² - 25r - 4r + 10 = 0 

⇒ 5r(2r - 5) - 2(2r - 5) = 0 

⇒ (2r - 5)(5r - 2) = 0

⇒ r = \(\frac{5}{2}, \frac{2}{5}\)

Therefore the first three terms are:

(i) if r = \(\frac{5}{2}\) then

\(\frac{2}{5}, 1 , \frac{5}{2}\)

(ii) if r = \(\frac{2}{5}\)then

\(\frac{5}{2},1,\frac{2}{5}\)

Hence, the Common ratio r = \(\frac{5}{2},\frac{2}{5}\) and the first three terms are:

(i) if r = \(\frac{5}{2}\) then

\(\frac{2}{5},1,\frac{5}{2}\)

(ii) If r = \(\frac{2}{5}\) then

\(\frac{5}{2},1,\frac{2}{5}\)

We have been given that 2^nd, 3^rd and 6^th terms of an AP are the three consecutive terms of a GP. 

Let the three consecutive terms of the G.P. be a,ar,ar² . 

Where a is the first consecutive term and r is the common ratio. 

2^nd, 3^rd terms of the A.P. are a and ar respectively as per the question. 

∴ The common difference of the A.P. = ar - a 

And the sixth term of the A.P. = ar² 

Since the second term is a and the sixth term is ar² (In A.P.) 

We use the formula: t = a + (n - 1)d 

∴ ar² = a + 4(ar - a)…(the difference between 2^nd and 6^th term is 4(ar - a)) 

⇒ ar² = a + 4ar - 4a 

⇒ ar² + 3a - 4ar = 0 

⇒ a(r² - 4r + 3) = 0 

⇒ a(r - 1)(r - 3) = 0 

Here, we have 3 possible options: 

1) a = 0 which is not expected because all the terms of A.P. and G.P. will be 0. 

2) r = 1,which is also not expected because all th terms would be equal to first term. 

3) r = 3,which is the required answer. 

Hence, Common ratio = 3

It is given that: 

a ^1/x = b^1/y = c^1/z 

Let a^1/x = b^1/y = c^1/z = k 

⇒ a ^1/x = k 

⇒ (a^1/x) ^x = k^x…(Taking power of x on both sides.) 

⇒ a ^{1/x × x} = k^x 

⇒ a = k^x 

Similarly b = k^y 

And c = k^z 

It is given that a,b,c are in G.P. 

⇒ b² = ac 

Substituting values of a,b,c calculated above, we get: 

⇒ (ky )² = k^xk ^z 

⇒ k²y = k^{x + z} 

Comparing the powers we get, 

2y = x + z 

Which is the required condition for x,y,z to be in A.P. 

Hence, proved that x,y,z, are in A.P.

Given: 

Sum of GP = 728 

∴ Common Ratio = r = \(\frac{6}{2}\) = 3

 a = 2

 To find: Number of terms = n. 

Sum of GP for n terms = \(\frac{a(r^n-1)}{r-1}\)

⇒ 728 = \(\cfrac{2(3^n -1)}{3-1}\)

⇒ 728 = 3ⁿ - 1 

⇒ 729 = 3ⁿ 

⇒ 3⁶ = 3ⁿ 

∴ n = 6.