Sequences And Series Of Real Numbers

(i) Here, a_n =

For n = 1, a₁ = = =

For n = 2, a₂ = = = = 0

For n = 3, a₃ = = = = 1

Hence, the first three terms of the sequence are, 0 and 1.

(ii) Here, c_n = (-1)ⁿ 3^{n + 2}

For n = 1, c₁ = (-1)¹ 3^{1 + 2} = (-1) 3³ = (-1) (27) = -27

For n = 2, c₂ = (-1)² 3^{2 + 2} = (1) 3⁴ = (1) (81) = 81

For n = 3, c₃ = (-1)³ 3^{3 + 2} = (-1) 3⁵ = (-1) (243) = -243

Hence, the first three terms of the sequence are -27, 81 and -243.

(iii) Here, z_n =

For n = 1, z₁ = = =

For n = 2, z₂ = = = = 2

For n = 3, z₃ = = =

Hence, the first three terms of the sequence are, 2 and.

(i) Here, a_n =

For n = 7, a₇ = = =

For n = 9, a₉ = = =

(ii) Here, a_n = (-1)ⁿ 2^{n + 3} (n + 1)

For n = 5, a₅ = (-1)⁵ 2^{5 + 3} (5 + 1) = (-1) 2⁸ (6) = -6 (256) = -1536

For n = 8, a₈ = (-1)⁸ 2^{8 + 3} (8 + 1) = (1) 2¹¹ (9) = 9 (2048)

= 18432

(iii) Here, a_n = 2n² – 3n + 1

For n = 5, a₅ = 2(5)² – 3(5) + 1 = 2(25) – 15 + 1 = 50 – 15 + 1

= 36

For n = 7, a₇ = 2(7)² – 3(7) + 1 = 2(49) – 15 + 1 = 98 – 21 + 1

= 78

(iv) Here, a_n = (-1)ⁿ (1 – n + n²)

For n = 5, a₅ = (-1)⁵ (1 – 5 + 5²) = (-1) (1 – 5 + 25) = (-1) (21)

= -21

For n = 8, a₅ = (-1)⁸ (1 – 8 + 8²) = (1) (1 – 8 + 64) = (1) (57)

= 57

For n = 18, n is even.

So, a₁₈ = 18(18 + 3) = 18(21) = 378

For n = 25, n is odd.

So, a₂₅ = = = =

For n = 13, n is odd.

So, b_n = 13(13 + 2) = 13(15) = 195

For n = 16, n is even.

So, b_n = 16² = 256

Given that a₁ = 2, a₂ = 3 + a₁ and a_n = 2a_n-1 + 5 for n > 2.

Now, a₁ = 2

⇒ a₂ = 3 + a₁ = 3 + 2 = 5

⇒ a₃ = 2a₂ + 5 = 2(5) + 5 = 10 + 5 = 15

⇒ a₄ = 2a₃ + 5 = 2(15) + 5 = 30 + 5 = 35

⇒ a₅ = 2a₄ + 5 = 2(35) + 5 = 70 + 5 = 75

∴ The required terms of sequence are 2, 5, 15, 35 and 75.

Given that a₁ = a₂ = a₃ = 1 and a_n = a_n-1 + a_n-2 for n > 3.

Now, a₁ = 1

⇒ a₂ = 1

⇒ a₃ = 1

⇒ a₄ = a₃ + a₂ = 1 + 1 = 2

⇒ a₅ = a₄ + a₃ = 2 + 1 = 3

⇒ a₆ = a₅ + a₄ = 3 + 2 = 5

∴ The required terms of the sequence are 1, 1, 1, 2, 3 and 5.

First term, a = 6; Common difference, d = 5

We know that the AP is in the form of a, a + d, a + 2d, a + 3d … a + (n – 1) d, a + nd …

⇒ AP = 6, 6 + 5, 6 + 2(5), 6 + 3(5) …

= 6, 11, 6 + 10, 6 + 15 …

= 6, 11, 16, 21 …

∴ The required A.P. is 6, 11, 16, 21 …

We know that the general form, t_n = a + (n – 1) d.

⇒ t_n = 6 + (n – 1) 5

= 6 + 5n – 5

= 5n + 1

∴ The required general term, t_n = 5n + 1

The given A.P. is 125, 120, 115, 110 …

We know that common difference, d = a₁ – a.

⇒ d = 120 – 125

= -5

∴ Common difference = -5

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₁₅ = 125 + (15 – 1) (-5)

= 125 + 14 (-5)

= 125 – 70

= 55

∴ 15^th term of A.P. = 55

Here, first term, a = 24

We know that common difference, d = a₁ – a.

We know that nth term of A.P. , t_n = a + (n – 1) d.

To find the nth term, here t_n = 3

⇒ 3 = 24 + (n – 1) (-3/4)

⇒ 3 – 24 = (n – 1) (-3/4)

⇒ -21 = (n – 1) (-3/4)

⇒ -21 × = n – 1

⇒ 28 = n – 1

⇒ n = 28 + 1 = 29

∴ t₂₉ of the given A.P. is 3.

Given A.P. √2, 3√2, 5√2 …

Here, first term, a = √2

We know that common difference, d = a₁ – a.

∴ d = 3√2 - √2 = 2√2

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₁₂ = √2 + (12 – 1) (2√2)

= √2 + 11 (2√2)

= √2 + 22 √2

= 23√2

∴ 12^th term of A.P. i.e. t₁₂ = 23√2

Given A.P. 4, 9, 14 …

Here, first term, a = 4

We know that common difference, d = a₁ – a.

∴ d = 9 - 4 = 5

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₁₇ = 4 + (17 – 1) (5)

= 4 + 16 (5)

= 4 + 80

= 84

∴ 17^th term of A.P. i.e. t₁₇ = 84

(i) Here, first term, a = -1

Last term, l =

We know that common difference, d = a₁ – a.

∴ d = – (-1) = + 1 = =

We know that Number of terms, n = () + 1.

⇒ n = + 1

= + 1

= + 1

= (13 × 2) + 1

= 26 + 1

= 27

∴ There are 27 terms in the given A.P.

(ii) Here, first term, a = 7

Last term, l = 205

We know that common difference, d = a₁ – a.

∴ d = 13 - 7 = 6

We know that Number of terms, n = () + 1.

⇒ n = + 1

= + 1

= 33 + 1

= 34

∴ There are 34 terms in the given A.P.

Here, t₉ = 0

We have to prove that t₂₉ = 2t₁₉

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₉ = a + (9 – 1) d = a + 8d

But t₉ = 0

⇒ a + 8d = 0

∴ a = -8d … (i)

Now, t₂₉ = a + (29 – 1) d

= a + 28d

= -8d + 28d [From (i)]

= 20d

∴ t₂₉ = 20d … (1)

Now, t₁₉ = a + (19 – 1) d

= a + 18d

= -8d + 18d [From (i)]

= 10d

∴ t₁₉ = 10d … (2)

Equating (1) and (2),

⇒ 20d = 10d

⇒ 20d = 10d (2)

∴ t₂₉ = 2(t₁₉)

Hence proved.

Here, t₁₀ = 41 and t₁₈ = 73

We know that nth term of A.P. , t_n = a + (n – 1) d.

First, t₁₀ = a + (10 – 1) d = 41

⇒ a + 9d = 41 … (1)

Then, t₁₈ = a + (18 – 1) d = 73

⇒ a + 17d = 73 … (2)

From (1) and (2),

a + 9d = 41

a + 17d = 73

(-) (-) (-)

-8d = -32

⇒ d = 32/8 = 4

Substituting d = 4 in (1),

⇒ a + 9(4) = 41

⇒ a = 41 – 36 = 5

Now, t₂₇ = 5 + (27 – 1) (4)

= 5 + 26(4)

= 5 + 104

= 109

∴ The 27^th term i.e. t₂₇ = 109

For first A.P. , first term, a = 1; common difference, d = 7 – 1 = 6

For second A.P, first term, a = 100; common difference , d = 95 – 100 = -5

We know that nth term of A.P. , t_n = a + (n – 1) d.

First A.P:

⇒ t_n1 = 1 + (n – 1) (6)

= 1 + 6n – 6

∴ t_n1 = 6n – 5

Second A.P:

⇒ t_n2 = 100 + (n – 1) (-5)

= 100 – 5n + 5

∴ t_n2 = 105 – 5n

Given that nth terms of the two A.P.’s are the same.

∴ t_n1 = t_n2

⇒ 6n – 5 = 105 – 5n

⇒ 6n + 5n = 105 + 5

⇒ 11n = 110

⇒ n = 110/11 = 10

∴ For the nth terms of the two given A.P.’s to be same, the value of n = 10.

The two digits form the A.P.: 10, 11, 12 … 99.

The two digit numbers that are divisible by 13 form the A.P.: 13, 26, 39 … 91.

Here, first term, a = 13

We know that common difference, d = a₁ – a.

∴ d = 26 – 13 = 13

And last term, l = 91

We know that l = a + (n – 1) d.

⇒ 91 = 13 + (n – 1) (13)

⇒ 91 – 13 = (n – 1) (13)

⇒ 78 = (n – 1) (13)

⇒ 78/13 = (n – 1)

⇒ 6 = n – 1

⇒ n = 6 + 1 = 7

∴ There are 7 two digit numbers that are divisible by 13.

Here, t₇ = 1000 and t₁₀ = 1450

We know that nth term of A.P. , t_n = a + (n – 1) d.

First, t₇ = a + (7 – 1) d = 1000

⇒ a + 6d = 1000 … (1)

Then, t₁₀ = a + (10 – 1) d = 1450

⇒ a + 9d = 1450 … (2)

From (1) and (2),

a + 6d = 1000

a + 9d = 1450

(-) (-) (-)

-3d = -450

⇒ d = 450/3 = 150

Substituting d = 150 in (1),

⇒ a + 6(150) = 1000

⇒ a = 1000 – 900 = 100

Now, t₁ = 100 + (1 – 1) (150)

= 100 + 0

= 100

And t₁₅ = 100 + (15 – 1) (150)

= 100 + 14 (150)

= 100 + 2100

= 2200

∴ Number of TVs produced in the first year are 100 and in the 15^th year are 2200.

The savings form the A.P.: 640, 720, 800 …

Here, first term, a = 640

We know that common difference, d = a₁ – a.

∴ d = 720 – 640 = 80

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₂₅ = 640 + (25 – 1) (80)

= 640 + (24) (80)

= 640 + 1920

= 2560

∴ The man’s savings in the 25^th month are Rs. 2560.

We know that the three consecutive terms of an A.P. may be taken as a – d, a, a + d.

Given, a – d + a + a + d = 6

⇒ 3a = 6

⇒ a = 6/3 = 2

Also given (a – d) (a) (a + d) = -120

We know that (a – b) (a + b) = a² – b²

⇒ (a² – d²) (2) = -120

⇒ (2² – d²) = -120/2

⇒ 4 – d² = -60

⇒ d² = 60 + 4 = 64

⇒ d = 8

So, the numbers are

⇒ a – d = 2 – 8 = -6

⇒ a = 2

⇒ a + d = 2 + 8 = 10

∴ The three consecutive numbers are -6, 2, 10.

We know that the three consecutive terms of an A.P. may be taken as a – d, a, a + d.

Given, a – d + a + a + d = 18

⇒ 3a = 18

⇒ a = 18/3 = 6

Also given (a – d)² + a² + (a + d)² = 140

We know that (a – b)² = a² – 2ab + b²

And (a + b)² = a² + 2ab + b²

⇒ a² – 2ad + d² + a² + a² + 2ad + d² = 140

⇒ 3a² + 2d² = 140

⇒ 3(6)² + 2d² = 140

⇒ 108 + 2d² = 140

⇒ 2d² = 140 – 108 = 32

⇒ d² = 32/2 = 16

⇒ d = 4

So, the numbers are

⇒ a – d = 6 – 4 = 2

⇒ a = 6

⇒ a + d = 6 + 4 = 10

∴ The three consecutive numbers are 2, 6, 10.

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ First, t_n = a + (n – 1) d

⇒ Then, t_m = a + (m – 1) d

Given, mt_m = nt_n

⇒ m [a + (m – 1) d] = n [a + (n – 1) d]

⇒ ma + m²d – md = na + n²d – nd

⇒ ma – na + m²d – n²d – md + nd = 0

⇒ a (m – n) + d (m² – n²) – d (m – n) = 0

We know that a² – b² = (a – b) (a + b)

⇒ (m – n) [a + d (m + n) – d] = 0

⇒ [a + d (m + n) – d] = 0

⇒ a + (m + n – 1) d = 0

∴ (m + n)th term, t_{m + n} = 0

Hence proved.

Yes, the amounts form an A.P.

Given principal, p = Rs. 25,000

Simple Interest, r = 14%

Time, t = 20 years

We know that Total amount = p (1 + )

⇒ Total Amount = 25000 (1 + )

= 25000 (1 + )

= 25000 ()

= 95000

∴ Amount of investment after 20 years = Rs. 95, 000

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

We know that when t₁, t₂, t₃ … are in A.P., t₃ – t₂ = t₂ – t₁

⇒ c – b = b – a

⇒ 2b = a + c

Squaring on both sides,

⇒ (2b)² = (a + c)²

We know that (a + b)² = a² + 2ab + b².

⇒ 4b² = a² + 2ac + c²

Subtracting 4ac on both sides,

⇒ 4b² – 4ac = a² + 2ac + c²– 4ac

⇒ 4 (b² – ac) = a² – 2ac + c²

∴ 4 (b² – ac) = (a – c)²

Hence proved.

Given, a, b, c are in A.P.

Here, first term = a

Common difference, d₁ = b – a … (1)

and d₂ = c – b … (2)

Consider, and,

Common difference, d₃ = -

=

= … (3)

⇒ d₄ = -

=

= … (4)

From (1) and (2),

⇒ d₁ = d₂

⇒ b – a = c – b

Dividing both sides by abc,

⇒ =

⇒ d₃ = d₄ [From (3) and (4)]

Hence,, and are in A.P.

Given, a², b² and c² are in A.P.

We know that when t₁, t₂, t₃ … are in A.P., t₃ – t₂ = t₂ – t₁

⇒ b² – a² = c² – b²

We know that a² – b² = (a – b) (a + b)

⇒ (b – a) (b + a) = (c – b) (c + b)

⇒ =

Dividing by (c + a) on both sides,

⇒ =

⇒ =

⇒

Hence, , and are in A.P.

Let a^x = b^y = c^z = k.

We know that if a^m = k, then a =.

⇒ a = , b = , c =

Given, b² = ac

⇒ = ×

We know that (a^m)ⁿ = a^mn and a^m × aⁿ = a^{m + n}.

⇒ =

Bases are same, so we equate the powers.

⇒ = +

⇒ + = +

⇒ - = -

We know that when t₁, t₂, t₃ … are in A.P., t₃ – t₂ = t₂ – t₁

∴, and are in A.P.

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given: a₁ = 0.12, a₂ = 0.24, a₃ = 0.48

Now,

And,

Therefore,

Now, ∵ the ration is same,

⇒ 0.12, 0.24, 0.48, … is a G.P with common ratio = 2

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given: a₁ = 0.004, a₂ = 0.02, a₃ = 0.1

Now,

And,

Therefore,

Now, ∵ the ration is same,

⇒ 0.004, 0.02, 0.1, … is a G.P with common ratio = 5

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given:

Now,

And,

And,

Therefore,

Now, ∵ the ration is same,

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given: a₁ = 12, a₂ = 1,

Now,

And

Therefore,

Now, ∵ the ration is same,

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given:

Now,

And

Thherefore,

Now, ∵ the ration is same,

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given: a₁ = 4, a₂ = -2, a₃ = -1,

Now,

And,

And,

Therefore,

Now, ∵ the ration is not same,

Given:

; (where, n = 1,2,……n-1)

Now, taking n = 3.

⇒ r = -2

Also, a_n = a₁r^n-1 (n = no. of term and a₁ = first term of G.P)

∴ 10^th term (a₁₀ = a₁r⁹)

⇒ a₁₀ = -2⁷

⇒ common ratio = -2 and 10^th term = -2⁷

a₄ = 54 and a₇ = 1458

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒ a₄ = a₁r³

⇒ 54 = a₁r³ ………(1)

Also, ⇒ a₇ = a₁r⁶

⇒ 1458 = a₁r⁶ ………(2)

Dividing equation (1) & (2), we get-

⇒ r³ = 27

⇒ r³ = 3³

⇒ r = 3

Now, putting value of r in equation (1), we get-

⇒54 = a₁3³

⇒ a₁ = 2

Now, G.P will be-

⇒ a₁ , a₁r ,a₁r², a₁r³,………

⇒ 2, 2 × 3, 2 × 3², 2 × 3…

⇒ 2, 6, 18, 54,… is the G.P.

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒ a₆ = a₁r⁵

………(1)

Now, G.P will be-

⇒ a₁, a₁r, a₁r², a₁r³,………

(i) a₁ = 5, a₂ = 2

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒ n-1 = 7

⇒ n = 8

(ii) a₁ = 1, a₂ = 2

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒1024 = 1 × 2^n-1

⇒ 2¹⁰ = 2^n-1

⇒ n-1 = 10

⇒ n = 11

∴ 1024 is the 11^th term of G.P.

Given:For 1^st G.P-

a₁ = 162 and a₂ = 54,

And for 2^nd G.P-

.

R = 3

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

And ∵ nth term of both G.P are equal,

⇒ a_n = A_n.

⇒ a₁r^n-1 = A₁R^n-1

⇒ n-1 = 4

⇒ n = 5.

a₁ = 3 and a₅ = 1875

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒ a₅ = a₁r⁴

⇒1875 = 3r⁴ ………(1)

⇒ r⁵ = 625

⇒ r⁵ = 5⁵

⇒ r = 5 is the common ratio.

⇒ second term = a and third term = ar.

(where, r is common ratio)

……..(1)

Also, their product is 1.

⇒ a³ = 1

⇒ a = 1

Substituting a = 1 in equation (1)

⇒ 10 + 10r + 10r² = 39r

⇒ 10r²-29r + 10 = 0

⇒ 10r²-25r-4r + 10 = 0

⇒ 5r(2r-5)-2(2r-5)

⇒ (5r-2) (2r-5) = 0

Now, G.P will be-

Let the first term of G.P be .

⇒ second term = a and third term = ar.

(where, r is the common ratio)

∵ sum of three terms is 156

⇒ a³ = 216

⇒ a = 6. ……..(1)

Also,sum of their product in pairs is 216.

………(2)

Substituting (1) in (2), we get-

⇒ 36(1 + r + r²) = 156r

⇒ 36 + 36r + 36r² = 156r

⇒ 36 -120r + 36r² = 0

⇒ 12(3r²-10r + 3) = 0

⇒ 3r²-10r + 3 = 0

⇒ 3r²-9r-1r + 3 = 0

⇒ 3r(r-3) -1(r-3) = 0

⇒ (3r-1)(r-3) = 0

Now, G.P will be-

⇒18, 6, 2 or 2, 6, 18 are the three consecutive terms.

⇒ second term = a and third term = ar.

(where, r is the common ratio)

∵ sum of three terms is 7

……..(1)

………(2)

Now, ∵ R.H.S of equation (1) & (2) is equal-

⇒ L.H.S of equation (1) = L.H.S of equation (2)

⇒ a² = 4

⇒ a = √4

⇒ a = 2

Now substituting a = 1 in (2), we get-

⇒ 2 + 2r + 2r² = 7r

⇒ 2r²-5r + 2 = 0

⇒ 2r²-4r-r + 2 = 0

⇒ (2r-1)(r-2) = 0

Now, G.P will be-

⇒ 4, 2, 1 or 1, 2, 4 are the three consecutive terms.

Let the first term of G.P be a

⇒ second term = ar and third term = ar².

(where, r is the common ratio)

∵ sum of three terms is 13

⇒ a(1 + r + r²) = 13r……..(1)

Also, sum of their squares is 91.

⇒ a² (1 + r² + r⁴) = 91r² ………(2)

Now, Squaring (1) dividing by (2)

⇒ 7( 1 + r² + r) = 13( 1 + r² - r)

⇒ (7 + 7r² + 7r) = ( 13 + 13r² - 13 r)

⇒ 6r² - 20r + 6 = 0

⇒ 2(3r² - 10r + 3) = 0

⇒ 3r² - 10r + 3 = 0

⇒ 3r² - 9r-r + 3 = 0

⇒ (3r – 1)(r – 3) = 0

Substituting r in equation (1), we get-

⇒a(1 + 3 + 9) = 13 × 3

And,

⇒13a = 13

And

⇒ a = 1 and a = 9.

Now, G.P is-

a, ar, ar²,….

⇒ If r = 3 and a = 1 then,

⇒1, 1×3, 1×3², ……

= 1, 3 , 9 are the first three terms.

And, If and a = 9 then,

= 9, 3, 1 are the first three terms.

⇒ 1,3,9,… … or 9,3,1,… … is the G.P.

∵ there is annual compounding interest

⇒ consecutive amounts are forming G.P.

Now,a₁₂ = ?

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

∵ there is annual depreciation at a constant rate per year

⇒ consecutive value are forming G.P.

Now,a₁₅ = ?

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

Proof:∵ a, b, c, d are in G.P

⇒ a = a, b = ar, c = ar²,d = ar³.

⇒ L.H.S = (a - ar + ar²)(ar + ar² + ar³)

= a²r (1 + r² + r⁴ )

And, R.H.S = a²r + a²r³ + a²r⁵

= a²r (1 + r² + r⁴ )

⇒ L.H.S = R.H.S

Hence, proved that-

(a - b + c) (b + c + d) = ab + bc + cd.

Proof: ∵ a, b, c, d are in G.P

⇒ a = a, b = ar ,c = ar²,d = ar³.

To prove: a + b, b + c, c + d, are also in G.P, if-

⇒ (a + b) (c + d) = (b + c)²

Now, we need to prove : (a + b) (c + d) = (b + c)²

L.H.S. = (a + ar)(ar² + ar³)

= a(1 + r) ar² (1 + r)

= a²r² (1 + r)²

R.H.S. = (ar + ar²)²

= (ar(1 + r))²

= a²r² (1 + r)²

⇒ L.H.S = R.H.S

Hence, proved that-

(a + b) (c + d) = (b + c)²

⇒ a + b, b + c, c + d, are also in G.P

(i) In the A.P.

First term = 1

No. of terms = 75

Common difference = 1

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 2850

(ii) In the A.P.

First term = 1

No. of terms = 125

Common difference = 1

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 7875

In the A.P.

First term = 3 + 2×1 = 5

No. of terms = 30

Last term = 3 + 2×30 = 63

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 1020

(i) In the A.P.

First term = 38

Last term = 2

Common difference = 35–38 = –3

Nth term = a + (n–1) d

⇒ 2 = 38 + (n–1)(–3)

⇒ –36 = (n–1)(–3)

⇒ 12 = (n–1)

⇒ n = 13

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 260

(ii) In the A.P.

First term = 6

No. of terms = 25

Common difference = 21/4 – 6 = –3/4

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = –75

(i) Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 1890

(ii) Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 50

Let the sum of n terms be,

S_n= 1² –2² +3² –4²+5² –6²+7² –8² ….

= (1² –2²) + (3² –4²) + (5² –6²) + (7²–8²) …..

= (1–4) + (9–16) + (25–36) + (49–64)……..

= –3 –7 –11 –15………………. [No. of terms ]

This represents an A.P.

(NOTE: Here the number of terms has been halved. If n= no. of terms in original series and m= no. of terms in new A.P. then )

Now, here a= –3, d= (–7 – (–3)) = –4

Sum of m terms =

=

=

=

=

=

As n= 40,

Required sum =

= –820

Sum of terms =

⇒ 44 =

⇒ 8 = 2a + 10d

⇒ 2a + 10d = 8 ……………………….(1)

Now sum of next 11 terms is 55

Sum of first 22 terms = Sum of first 11 terms + Sum of next 11 terms

⇒ Sum of first 22 terms = 44 + 55 = 99

Sum of terms =

⇒ 99 =

⇒ 9 = 2a + 21d

⇒ 2a + 21d = 9 ……….(1)

Subtracting (1) from (2), we get

⇒ 11d = 1

⇒ d = 1/11

Putting value of d in (1), we get

⇒ 2a + 10(1/11) = 8

⇒ 2a = 8–(10/11)

⇒ 2a = 78/11

⇒ a = 39/11

Therefore, the series is

In the A.P.

First term = 60

Common difference = 56 – 60 = –4

Sum of terms =

⇒ 368 =

⇒ 368 =

⇒ 736 = n (124 – 4n)

⇒ 4n² – 124n + 736 = 0

⇒ 4n² – 92n–32n + 736 = 0

⇒ 4n(n–23) –32(n–23) = 0

⇒ (4n–32)(n–23) = 0

⇒ 4n = 32 or n = 23

⇒ n = 8 or n = 23

Therefore, no. of terms can be 8 or 23

Series of three digit numbers divisible by 9 is:

108, 117,………………………………999

In the A.P.

First term = 108

Last term = 999

Common difference = 9

Nth term = a + (n–1) d

⇒ 999 = 108 + (n–1)9

⇒ 891 = (n–1)9

⇒ 99 = (n–1)

⇒ n = 100

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 55350

First term = a

Common difference = d

3^rd term = a + (3–1)d = a + 2d

7^th term = a + (7–1)d = a + 6d

Now, 3^rd term = 7

⇒ a + 2d = 7 ………………………(1)

And, 7^th term = 3 × (3^rd term) + 2

⇒ a + 6d = 3(a + 2d) + 2

⇒ a + 6d = 3a + 6d + 2

⇒ 2a = –2

⇒ a = –1

Putting value of a in (1)

(⇒ –1) + 2d = 7

⇒ 2d = 8

⇒ d = 4

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 740

Series of all natural numbers divisible by 11 between 300and 500 is:

308, 319,………………………………495

In the A.P.

First term = 308

Last term = 495

Common difference = 11

Nth term = a + (n–1) d

⇒ 495 = 308 + (n–1)11

⇒ 187 = (n–1)11

⇒ 17 = (n–1)

⇒ n = 18

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 7227

In the A.P.

First term = 1

Common difference = 6 – 1 = 5

Sum of terms =

⇒ 148 =

⇒ 148 =

⇒ 296 = n (5n–3)

⇒ 5n² – 3n –296 = 0

⇒ 5n² –40n + 37n –296 = 0

⇒ 5n(n–8) + 37(n–8) = 0

⇒ (n–8)(5n + 37) = 0

⇒ n = 8 or 5n = –37

As negative value of n is not possible, n = 8

⇒ x = 8^th term of the series

⇒ x = 1 + (8–1)5

⇒ x = 1 + 35 = 36

We first find the sum of all numbers divisible by 5

Series of all natural numbers divisible by 5 between 100and 200 is:

105, 110,………………………………195

In the A.P.

First term = 105

Last term = 195

Common difference = 5

Nth term = a + (n–1) d

⇒ 195 = 105 + (n–1)5

⇒ 90 = (n–1)5

⇒ 18 = (n–1)

⇒ n = 19

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 2850

Sum of 101,102….199 = Sum of 199 natural numbers – Sum of 100 natural numbers

⇒ Sum(101, 102,…. 199) =

⇒ Sum(101, 102,…. 199) = 19900–5050 = 14850

Sum of all numbers between 100 and 200 not divisible by 5 = 14850–2850 = 12000

The amount to be paid first paid = ₹4000

It increases each day by = ₹1000

Total amount that can be paid = ₹165000

a = 4000, d = 1000, Sum = 165000

Sum of terms =

⇒ 165000 =

⇒ 165 =

⇒ 330 = n (n + 7)

⇒ n² + 7n –330 = 0

⇒ n² –15n + 22n –330 = 0

⇒ n(n–15) + 22(n–15) = 0

⇒ (n–15)(n + 22) = 0

⇒ n = 15 or n = –22

As negative value of n is not possible, n = 15

Therefore, number of days that construction can be delayed = 15days.

Simple interest = P* R * T /100

P = ₹1000

R = 8

T = 1

Simple interest = 1000*8*1/100 = 80

Interest at the end of first year = ₹80

For second year,

P = 1000 + 1000 = 2000

R = 8

T = 1

Simple interest = 2000*8*1/100 = 160

Interest at the end of second year = ₹160

Similarly, for third year, P = 1000 + 1000 + 1000 = 3000

SI = 240.

Yes the simple interests form an A.P.

a = 80

d = 160–80 = 80

n = 30

Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 37200

For n = 1,

Sum = 3(1)²–2(1) = 1

Therefore, first term = 1

For n = 2,

Sum = 3(2)² – 2(2) = 12 – 4 = 8

Second term = 8–1 = 7

For n = 3,

Sum = 3(3)² – 2(3) = 21

Third term = 21– 8 = 13

Series : 1, 7, 13…..

This is an arithmetic progression as the difference between two terms is constant.

Common difference = 7–1 = 13–7 = 6

The clock strikes once at 1, twice at 2 ….so it strikes 12times at 12

In a day this striking from 1 to 12 happens twice.

No. of strikes in one turn from 1 to 12 = 1 + 2 + 3 + …..12

⇒ No. of strikes in one turn from 1 to 12 =

As the striking happens twice in a day,

Total number of strikes in a day = 78 × 2 = 156

First term = a

Common difference = (b–a)

Last term = c

⇒ c = a + (n–1)(b–a)

⇒ (n–1) = (c–a) / (b–a)

⇒ n = (b + c – 2a) /( b – a)

Sum of terms =

⇒ Sum of terms =

Hence proved.

In the A.P, let

First term = a

Common difference = d

Number of terms = (2n + 1)

Series: a,a + d,a + 2d……a + 2nd

For Odd terms

: a, a + 2d,…a + 2nd

First term = a

Common difference = 2d

Number of terms = n + 1

Sum of terms =

Sum of odd terms =

⇒ Sum of odd terms =

For Even terms

: a + d, a + 3d,…a + (2n–1)d

First term = a + d

Common difference = 2d

Number of terms = n

Sum of terms =

Sum of even terms =

⇒ Sum of even terms =

Sum of odd terms : Sum of even terms = : =

∴ Sum of odd terms : Sum of even terms = (n + 1) : n

Sum of m terms =

Sum of n terms =

Sum of m terms : Sum of n terms = m² : n²

⇒ : = m² : n²

⇒ n²m(2a + (m–1)d) = nm²( 2a + ( n–1)d)

⇒ 2an²m + n²m²d – n²md = 2anm² + n²m²d – nm²d

⇒ 2anm(n–m) = nmd(n–m)

⇒ 2a = d

mth term : nth term = a + (m–1)d : a + (n–1)d

⇒ mth term : nth term = a + (m–1)2a : a + (n–1) 2a

⇒ mth term : nth term = (2m–1) : (2n–1)

Longest side = 97

Each row decreases by = 2×2 = 4

Number of rows = 25

a = 97, d = –4, n = 25

Sum of n terms =

⇒ Sum of terms =

⇒ Sum of terms =

⇒ Sum of terms = 1225

In the G.P.,

First term = 5/2

Common ratio = =

Sum of n terms =

⇒ Sum of 20 terms =

⇒ Sum of 20 terms =

In the G.P.,

First term = 1/9

Common ratio = =

Sum of n terms =

⇒ Sum of 27 terms =

⇒ Sum of 27 terms =

(i) t₈ = 384

⇒ ar^8–1 = 384

⇒ (3)r⁷ = 384

⇒ r⁷ = 128

⇒ r = 2

Sum of n terms =

⇒ Sum of 8 terms =

⇒ Sum of 8 terms = 3(255) = 765

(ii) Sum of n terms =

⇒ Sum of 12 terms =

⇒ Sum of 12 terms =

(i) In the G.P.,

First term = 1

Common ratio = = 0.1

Sum of n terms =

⇒ Sum of 10 terms =

⇒ Sum of 10 terms =

(ii) Series = 1 + 11 + 111 + …..20 terms

Series = [9×(1 + 11 + 111 + …..)] ( Multiplying and dividing by 9)

⇒ Series = [9 + 99 + 999 + …..]

⇒ Series = [(10–1) + (100–1) + (1000–1) + …..]

⇒ Series = [10 + 100 + 1000 + ….. – (20×1)]

⇒ Series = [10 + 100 + 1000 + …..] –

We find the sum of 10 + 100 + 1000…..20terms as:

First term = 10

Common ratio = = 10

Sum of n terms =

⇒ Sum of 20 terms =

⇒ Sum of 20 terms =

⇒ Series = [] –

⇒ Series = –

(i) First term = 3

Common ratio = = 3

Sum of n terms =

⇒ Sum of n terms =

⇒ 1092 =

⇒ 728 =

⇒ 729 =

⇒ n = 6

(ii) First term = 2

Common ratio = = 3

Sum of n terms =

⇒ 728 =

⇒ 728 =

⇒ 729 =

⇒ n = 6

⇒

⇒

Sum of n terms =

⇒ Sum of 23 terms =

⇒ Sum of 23 terms =

Let the first term = a

And common ratio = r

Series : a, ar, ar², ar³

⇒ a + ar = 9

⇒ a(1 + r) = 9 ………………(1)

⇒ ar² + ar³ = 36

⇒ ar²(1 + r) = 36

⇒ 9r² = 36 ( From Equation (1))

⇒ r² = 4

⇒ r = 2

Putting the value of r in (1),

⇒ a(1 + 2) = 9

⇒ 3a = 9

⇒ a = 3

∴ Series : 3 + 6 + 12 + 24

(i) Series = 7 + 77 + 777 + …..n terms

Series = [9×(1 + 11 + 111 + …..)] ( Multiplying and dividing by 9)

⇒ Series = [9 + 99 + 999 + …..]

⇒ Series = [(10–1) + (100–1) + (1000–1) + …..]

⇒ Series = [10 + 100 + 1000 + ….. – (n×1)]

⇒ Series = [10 + 100 + 1000 + …..] –

We find the sum of 10 + 100 + 1000…..n terms as:

First term = 10

Common ratio = = 10

Sum of n terms =

⇒ Sum of n terms =

⇒ Sum of n terms =

⇒ Series = [] –

⇒ Series = –

(ii) Series = 0.4 + 0.94 + 0.994 + …..n terms

Series = (1–0.6) + (1–0.06) + (1–0.006) + ……..

⇒ Series = n×1 – (0.6 + 0.06 + 0.006 + ….)

We find the sum of 0.6 + 0.06 + 0.006…..n terms as:

First term = 0.6

Common ratio = =

Sum of n terms =

⇒ Sum of n terms =

⇒ Sum of n terms =

⇒ Series = n –

People infected in first week = 5

More people infected by each person = 4

Number of weeks = 15

a = 5, r = 4, n = 15

Sum of n terms =

⇒ Sum of 15 terms =

⇒ Sum of 15 terms =

So, Number of people infected by 15^th week =

In the second option,

Number of mangoes on first day = 1

Number of times of mangoes on subsequent day = 2

Number of days = 10

a = 1, r = 2, n = 10

Sum of n terms =

⇒ Sum of 10 terms =

⇒ Sum of 10 terms = = 1023

Therefore, the boy will get more mangoes in 2^nd case as there were only 1000 mangoes in the 1^st case.

In the G.P.,

Let First term = a,

Common ratio = r

Series: a, ar, ar²,…….ar^n–1

Sum of all terms =

For odd terms,

a, ar²,………ar^n–2

First term = a

Common ratio = r²

Number of terms = n/2

Sum of odd terms =

⇒ Sum of odd terms =

Now,

Sum of all terms = 3× Sum of odd terms

⇒ = 3 ×

⇒ (1–r2) = 3(1–r)

⇒ r2–3r + 2 = 0

⇒ r2–2r–r + 2 = 0

⇒ r(r–2)–1(r–2) = 0

⇒ (r–1) (r–2) = 0

r = 1 or r = 2

But r = 1 is not possible, So r = 2.

Sum of n terms =

S₁ =

S₂ =

S₃ =

Putting value of S₁, S₂ and S₃ on the left side, we get:

S₁(S₃ – S₂) =

⇒ S₁(S₃ – S₂) =

⇒ S₁(S₃ – S₂) =

⇒ S₁(S₃ – S₂) =

⇒ S₁(S₃ – S₂) = ………..(1)

Now, we solve the right side by putting S₁, S₂ and S₃ :
(S₂– S₁)² =

⇒ (S₂– S₁)² = ………….(2)

From (1) and (2), we have:

Left hand side = Right Hand side

Hence Proved.

Given that series S = 1 + 2 + 3 + 4 + 5… + 45, it has n = 45 terms and to find the sum S.

Formula for sum of first n numbers is

S =

45x23

= 1035

The sum S = 1 + 2 + 3 + … + 45 = 1035

Given the the series S = 16² + 17² + 18 + … + 25²

Let S₁ = 1² + 2² + 3² + … + 25² with n = 25

Let S₂ = 1² + 2² + 3² + .. + 15² with n = 15

Required sum S = S₁-S₂

To find the sum S₁ -

Formula to find the sum of first n squares of natural numbers is

S =

S₁ =

=

S₁ = 5525

To find the sum S₂ -

Formula to find the sum of first n squares of natural numbers is

S =

S₂ =

S₂ =

S₂ = 1240

Required sum S = S₁-S₂

S = 5525-1240 = 4285

The sum S = 16² + 17² + 18 + … + 25² = 4285

Given the series S = 2 + 4 + 6 + … + 100,

We see that there is a common term 2 in all the numbers in the series. Taking 2 common, we have

S = 2(1 + 2 + 3 + .. + 50)

Let 1 + 2 + 3 + .. + 50 be S₁, with n = 50

S = 2S₁

To find S₁ -

Formula for sum of first n numbers is

S₁ =

25x51

= 1275

S = 2S₁

= 2 × 1275 = 2550

S = 2550

The sum S = 2 + 4 + 6 + … + 100 = 2550

Given the series S = 7 + 14 + 21…. + 490,

We see that there is a common term 7 in all the numbers in the series. Taking 7 common, we have

S = 7(1 + 2 + 3 + .. + 70)

Let 1 + 2 + 3 + .. + 70 be S₁, with n = 70

So S becomes S = 7S₁

To find S₁-

Formula for sum of first n numbers is

S₁ =

35 × 71

= 2485

S = 7S₁

S = 7 × 2485 = 17395

The sum S = 7 + 14 + 21…. + 490 = 17395

Given the the series S = 5² + 7² + 9² + … + 39²

Let S₁ = 1² + 3²

Let S₂ = 2² + 4² + 6² + .. + 38² with n = 19

Let S₃ = 1² + 2² + 3² + 4² + .. + 39² with n = 39

Required sum S = S₃-S₂-S₁

To find the sum S₁ -

S₁ = 1 + 9 = 10

To find the sum S₂ –

S2 can be rewritten as S₂ = 2²(1² + 2² + 3² + 4² + .. + 19²)

Formula to find the sum of first n squares of natural numbers is

S =

S₂ =

=

=

S₂ = 9880

To find the sum S₃ -

Formula to find the sum of first n squares of natural numbers is

S =

S₃ =

=

= 20540

Required sum S = S₃-S₂-S₁

S = 20540-9880-10 = 10650

The sum S = 5² + 7² + 9² + … + 39² = 10650

Given the the series S = 16³ + 17³ + … + 35³

Let S₁ = 1³ + 2³ + 3³ + … + 35³ with n = 35

Let S₂ = 1³ + 2³ + 3³ + .. + 15³ with n = 15

Required sum S = S₁-S₂

To find the sum S₁ -

Formula to find the sum of first n cubes of natural numbers is

S =

S₁ =

=

S₁ = 396900

To find the sum S₂ -

Formula to find the sum of first n cubes of natural numbers is

S =

S₂ =

=

S₂ = 14400

Required sum S = S₁-S₂

S = 396900-14400 = 382500

The sum S = 16³ + 17³ + … + 35³ = 382500

Given the first k cubes of natural numbers and that their sum is 6084,

By formula we have S =

6084 =

Taking square root on both sides, we get = 78

k(k + 1) = 156

k² + k = 156

Or k² + k-156 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -156

We get,

k = is invalid because it yields a negative k which doesn’t make sense because number of terms in a series cannot be negative.

= 12

The sum S = 1³ + 2³ + 3³ + … + k³ = 6084 corresponds to k = 12.

Given the first k cubes of natural numbers and that their sum is 2025,

By formula we have S =

2025 =

Taking square root on both sides, we get = 45

k(k + 1) = 90

k² + k = 90

Or k² + k-90 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -90

We get,

k = is invalid because it yields a negative k which doesn’t make sense because number of terms in a series cannot be negative.

= 9

The sum S = 1³ + 2³ + 3³ + … + k³ = 2025 corresponds to k = 9.

Given that the series S = 1 + 2 + 3 + … + p = 171

We have S =

p(p + 1) = 342

p² + p = 342

Or p² + p-342 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -342

We get,

p = is invalid because it yields a negative p which doesn’t make sense because number of terms in a series cannot be negative.

= 18

S = 1³ + 2³ + 3³ + ... + p³ where p = 18

Formula to find the sum of first n cubes of natural numbers is

S =

S =

S =

S = 29241

The sum S = 1³ + 2³ + 3³ + ... + p³ corresponds to p = 18 and S = 29241.

Given that the series S = 1³ + 2³ + 3³ + … + k³ = 8281

Formula to find the sum of first k cubes of natural numbers is

S =

8281 =

Taking square root on both sides,

91 =

k(k + 1) = 182

k² + k = 182

Or k² + k-182 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -182

We get,

k = is invalid because it yields a negative k which doesn’t make sense because number of terms in a series cannot be negative.

= 13

The sum S = 1³ + 2³ + 3³ + ... + k³ corresponds to k = 13.

Given series 1 + 2 + 3…. + k, we have k = 13

We have S =

=

=

S = 91

The sum S = 1³ + 2³ + 3³ + ... + k³ corresponds to k = 13 and 1 + 2 + 3…. + k = 91.

Given that there are 12 squares,

We see that their sides 12cm, 13cm..,23cm are in series.

To find the total area, we know that the area of a square is simply l² where l is the length of the side.

Area of first square = 12cm × 12cm = 12² cm²

Area of the second square = 13cm × 13cm = 13² cm²

And so on.

We observe that this is in a series.

So S = 12² + 13² + 14² + … + 23²

To find S,

Let S₁ be 1² + 2² + …. + 23² with n = 23

Let S₂ be 1² + 2² + …. + 11² with n = 11

S = S₁-S₂

To find S₁

1² + 2² + …. + 23² with n = 23

Formula to find the sum of first n squares of natural numbers is

S =

S₁ =

=

S₁ = 4324

To find S₂

1² + 2² + …. + 11² with n = 11

Formula to find the sum of first n squares of natural numbers is

S =

S₂ =

=

S₂ = 506

S = S₁-S₂

S = 4324-506

S = 3818cm²

The required sum S = 12² + 13² + 14² + … + 23² = 3818cm²

Given that there are 15 cubes,

We see that their sides 16cm, 17cm..,30cm are in series.

To find the total volume, we know that the volume of a cube is simply l³ where l is the length of the side.

Volume of first cube = 16cm × 16cm × 16cm = 16³ cm³

Volume of second cube = 17cm × 17cm × 17cm = 17³ cm³

And so on.

We observe that this is in a series.

So S = 16³ + 17³ + 18³ + … + 30³

To find S,

Let S₁ be 1³ + 2³ + …. + 30³ with n = 30

Let S₂ be 1³ + 2³ + …. + 15³ with n = 15

S = S₁-S₂

To find S₁

1³ + 2³ + …. + 30³ with n = 30

Formula to find the sum of first n cubes of natural numbers is

S =

S₁ =

=

S₁ = 216225

To find the sum S₂ –

1³ + 2³ + …. + 15³ with n = 15

Formula to find the sum of first n cubes of natural numbers is

S =

S₂ =

=

S₂ = 14400

S = S₁-S₂

S = 216225-14400

S = 201825 cm³

The required sum S = 16³ + 17³ + 18³ + … + 30³ = 201825 cm³

“Not true” – tells us that there is one option among A, B C or D which is false, while the rest are true.

Let us examine each option separately.

● A, is true because a sequence a is defined on a set of natural numbers.

● C, is true because an infinite sequence is possible. (ex. Infinite GP)

● D, is true because a finite sequence is possible. (ex. Finite AP, GP or HP)

● B, is false because every function is not a sequence, but every sequence is a function.

Hence, the correct option is B.

The above series is also called a Fibonacci sequence, where each term is the sum of its preceding two terms.

a_n = a_n-1 + a_n-2

We are interested in 8^th term so substituting n = 8 in the above equation,

a₈ = a₇ + a₆. Also, a₇ = a₆ + a₅

a₈ = (a₆ + a₅) + a₆

a₈ = (8 + 5) + 8

a₈ = 21

So the correct option is D.

The general term would be

Where n = 1,2,3….

Next term after is the fifth term so n = 5

We have

=

Therefore, the correct option is C.

If a, b, c, l, m are in A.P, then

b-a = c-b = l-c = m-l.

a-4b + 6c-4l + m becomes

a-b-b-b-b + c + c + c + c + c + c-l-l-l-l + m

Grouping the terms

We have

a + 4(c-b)-2(l-c)-2l + m

a + 2(c-b)-2(l-c) + 2(c-b)-2l + m

Since c-b = l-c,

a + 2(c-b)-2l + m

a + 2c-2b-2l + m

-(b-a) + (c-b)-(l-c) + (m-l)

= 0

So, the correct option is D.

If a,b,c are in AP, then b-a = c-b

Or

Or

Therefore, the correct option is D.

given nth term = 100n + 10

This can be rewritten as 110 + (n-1)100

This is in the form T_n = a + (n-1)d which forms an AP.

Therefore, the correct option is A.

Given

Let common ratio be d

a₄ = a₁ + 3d

a₇ = a₁ + 6d

2a₁ + 6d = 3a₁ + 18d

a₁ + 12d = 0

Which corresponds to the 13^th term (a₁ + (13-1)d) = 0

Therefore, the correct option is B.

Terms collected from an AP with a common interval between two terms is always in AP, a₅, a₁₀, a_15…. have a common interval of 5, so it is also in AP.

From AP, it is clear that

(4k-6)-(k + 2) = (3k-2)-(4k-6)

3k-8 = -k + 4

4k = 12 or k = 3.

So the correct choice is B.

Terms of an AP if multiplied with a constant will still be in AP.

The constant term is 3.

Therefore, 3a, 3b, 3c…are in AP.

Adding a constant to an AP does not change the sequence type. i.e. It will still be in AP.

The constant term added is 7.

Therefore, 3a + 7, 3b + 7….also form an AP.

The correct choice is B.

2 = ar^n-1 = ar^3-1 = ar²

Product of the first five terms = axarxar²xar³xar⁴

This can be rewritten as a⁵r¹⁰ or (ar²)⁵

So, Product of the first five terms is 2⁵

Hence the correct option is B.

Given that a, b, c are in GP,

Subtracting 1 from both sides, we get

Therefore, the correct option is A.

Any GP if multiplied with a constant term remains in GP.

x, 2x + 2, 3x + 3 are in GP, 5(x), 5(2x + 2), 5(3x + 3) will also form a GP because a constant term 5 is merely multiplied.

So the correct option is B.

The given sequence is -3, -3, -3

Test for AP

If a,b,c…are in AP the b-a = c-b…

-3-(-3) = -3-(-3) = 0

Therefore, the series is in AP.

Test for GP

If a,b,c.. are in GP then …

Therefore, the series is in GP.

The series is both in AP and GP.

So the correct option is D.

Given the product of first four consecutive terms of GP = 256.

Let the first four terms be a, ar, ar², ar³

Product = 256 = axarxar²xar³ = a⁴r⁶

Also given r = 4,

We have 256 = a⁴ × (4)⁶

a⁴ =

a = which is positive.

Third term is ar² = = 8

So the correct option is A.

Common ratio = =

So the correct option is B.

We see that the problem under consideration is a GP with a = 1 and common ratio r = sec x.

Sum of GP S =

Where n = number of terms, here n = 6.

Therefore

S =

The numerator is of the form a⁶-b⁶

a⁶-b⁶ = (a + b)(a-b)(a² + b² + ab)(a² + b²-ab)

Therefore (sec x)⁶ -1⁶ = (sec x + 1)(sec x-1)(sec²x + 1 + sec x)(sec²x + 1-sec x)

(sec x)⁶ -1⁶ = (sec x + 1)(sec x-1)(1 + sec²x + sec⁴x)

S = = (sec x + 1)(1 + sec²x + sec⁴x)

So the correct answer is option B.

Given t_n = 3-5_n, t₁ = 3-5 = -2

Sum of n terms of an AP = S_n = a + t_n)

S_n = -2 + 3-5n) = 1-5n)

So the correct answer is B.

If x, y, z are in GP, then Common ratio r = or

Common ratio in this problem is or = aⁿ

Therefore, the correct option is C.

Sum of first n natural numbers is S = = k

Sum of first n natural squares is S =

Which is k²

So the correct option is A.