Explore topic-wise InterviewSolutions in .

The correct answer is : (A) 45

Let the three consecutive terms in an A.P. be 

a – d, a and a + d. 

According to the first condition, 

a – d + a + a + d = 27 

∴ 3a = 27 

∴ a =  27/3

∴ a = 9 ….(i) 

According to the second condition, 

(a – d) a (a + d) = 504 

∴ a(a² – d² ) = 504 

∴ 9(a² – d² ) = 504 …[From (i)] 

∴ 9(81 – d² ) = 504 

∴ 81 – d² = 504

∴ 81 – d² = 56 

∴ d² = 81 – 56 

∴ d²= 25 

Taking square root of both sides, we get 

d = ± 5 

When d = 5 and a =9, 

a – d = 9 – 5 = 4 

a = 9 

a + d = 9 + 5 = 14 

When d = -5 and a = 9, 

a – d = 9 – (-5) = 9 + 5 = 14 

a = 9 

a + d = 9 – 5 = 4 

∴ The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.

For an A.P., let a be the first term and d be the common difference. 

S₅₅ =3300 …[Given] 

Since, S_n = n/2 [2a + (n – 1)d] 

∴ S₅₅ = 55/2 [2a + (55 – 1)d]

∴ 3300 = 55/2 (2a + 54d)

∴ 3300 = 55/2 x 2(a + 27d)

∴ 3300 = 55(a + 27d)

∴  a + 27d = 3300/5

∴ a + 27d = 60   ....(i)

Now, t_n = a + (n - 1)d

∴ t₂₈ = a + ( 28 - 1)d = a + 27d

∴ t₂₈ = 60   ... [Frtom (i)]

∴ 28^th terms of the A.P is 60.

4, 7, 10, 13, 16, ……

Correct option is A) 2, 4, 6, 8, 

(A) 2, 4, 6, 8, ......

\(a_2-a_1\) \(=4-2=2\)

\(a_3-a_2\) \(=6-4=2\)

and \(a_4-a_3\) \(=8-6=2\)

\(\because\) \(a_4-a_3\) \(=a_3-a_2\) \(=a_2-a_1\) \(=2\)

\(\therefore\) 2, 4, 6, 8, ...... is an arithmetic progression.

(B) 1, 2, 4, 8, ........

\(a_2-a_1\) \(=2-1=1,\)

\(a_3-a_2\) \(=4-2=2\)

\(\because\) \(a_3-a_2\) \(\neq a_2-a_1\)

\(\therefore\) 1, 2, 4, 8, ........ will not form an A.P.

(C) 4, 9, 16, 25, .......

\(a_2-a_1\) \(=9-4=5,\)

\(a_3-a_2\) \(=16-9=7\)

\(\because\) \(a_3-a_2\) \(\neq a_2-a_1\)

\(\therefore\) 4, 9, 16, 25, ....... will not form an A.P.

(D) 3, 9, 12, 18, .........

\(a_2-a_1\) \(=9-3=6,\)

\(a_3-a_2\) \(=12-9=3\)

\(\because\) \(a_3-a_2\) \(\neq a_2-a_1\)

\(\therefore\) 3, 9, 12, 18, ......... will not form an A.P.

Correct option is A) 2, 4, 6, 8, 

Let the required parts of 24 be (a - d),a and (a + d) such that they are in AP.

Then (a - d) + a + (a + d) = 24

\(\Rightarrow\) 3a = 24

\(\Rightarrow\) a = 8

Also, (a - d).a.(a + d) = 440

\(\Rightarrow\) a(a² - d²) = 440

\(\Rightarrow\) 8(64 - d²) = 440

\(\Rightarrow\) d² = 64 - 55 = 9

\(\Rightarrow\) d = \(\pm3\)

Thus, a = 8 and d = \(\pm3\)

Hence, the required parts of 24 are (5,8,11) or (11,8,5).

Properties of an Arithmetic Progression:

Property I:

For an A.P. with the first term ‘a’ and the common difference ‘d’, if any real number ‘k’ is added to each term, then the new sequence is also an A.P. with the first term ‘a + k’ and the same common difference ‘d’.

Property II:

For an A.P. with the first term ‘a’ and the common difference ‘d’, if each term of an A.P. is multiplied by any real number k, then the new sequence is also an A.P. with the first term ‘ak’ and the common difference ‘dk’.

Note:

1. If each term of an A.P. is multiplied by 0 then the new sequence will be 0, 0, 0, …

2. If each term of an A.P. is added, subtracted multiplied or divided by a certain constant then the new sequence is also an A.P.

Given:

l = a₉ = a + 8d = 28 and S₉ = 144 But, 

Now, S_n = \(\frac{n}{2}\)(a + l) 

144 = \(\frac{9}{2}\)(a + 28) 

⇒ 144 × \(\frac{2}{9}\) = a + 28

⇒ a + 28 = 32 

⇒ a = 4

Given : 

Sum of first three terms is 24 

To find : the first three terms of AP 

Assume the first three terms are a - d, a, a + d where a is the first term and d is the common difference 

So, 

Sum of first three terms is a - d + a + a + d = 24 

3a = 24 

a = 8 

Given that,

The product of three terms is 440 

So a³ - ad² = 440 

Substituting a = 8 

8³ - 8d² = 440 

512 - 8d² = 440 

72 = 8d² 

d = 3 or d = - 3 

Hence,

The given terms are a - d, a, a + d which is 5, 8, 11 or 11, 8, 5.

Let the four consecutive terms be a – 3d, a – d, a + d and a + 3d

 According to the first condition,

a – 3d + a – d + a + d + a + 3d = 12

\(\therefore\) 4a = 12    \(\therefore\) a = 12/4

\(\therefore\) a = 3 ... (i)

According to the second condition,

a + d + a + 3d = 14

\(\therefore\) 2a + 4d = 14

\(\therefore\) 2 x 3 + 4d = 14 ... [From (i)]

\(\therefore\) 4d = 14 – 6

\(\therefore\) 4d = 8 

\(\therefore\) d = 2

Thus, a –3d = 3 – 3 x 2 = –3

a – d = 3 – 2 = 1

a + d = 3 + 2 = 5

a + 3d = 3 + 3 x 2 = 9

\(\therefore\) The four consecutive terms of an A.P. are –3, 1, 5 and 9.

 We know that the sum of terms for different arithmetic progressions is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

51 terms of an AP whose a₂ = 2 and a₄ = 8

We know that, a₂ = a + d

2 = a + d  …(2)

Also, a₄ = a + 3d

8 = a + 3d  … (2)

Subtracting (1) from (2), we have

2d = 6

d = 3

Substituting d = 3 in (1), we get

2 = a + 3

⟹ a = -1

Given that the number of terms (n) = 51

First term (a) = -1

So,

S_n  = \(\frac{51}{2}\)[2(−1) + (51 − 1)(3)]

= \(\frac{51}{2}\)[−2 + 150]

= \(\frac{51}{2}\)[158]

= 3774

Hence, the sum of first 51 terms for the A.P. is 3774.

Given, 

First term, a₁ = 17 

Last term, a_n = 350 = l 

And difference, d = 9 

We know,

a_n = a + (n – 1)d 

350 = 17 + (n – 1) (9) 

350 = 8 + 9n 

342 = 9n 

n = 38

We know sum of n terms, S_n =  \(\frac{n}{2}\)[a + l]

S₃₈ = \(\frac{38}{2}\)[17 + 350]

= 19 x 367 = 6973

Hence, number of terms, n = 38

sum of n terms, S_n = 6973

Given A.P in which a = 17 

Last term = l = 350 

Common difference, d = 9 

We know that, a_n = a + (n – 1) d 

350 = 17 + (n- 1) 9 

⇒ 350 = 17 + 9n – 9 

⇒ 9n = 350 – 8 

⇒ n = \(\frac{342}{9}\) = 38 

Now, S_n = \(\frac{n}{2}\)(a + l) 

S₃₈ = \(\frac{38}{2}\)(17 + 350)

= 19 × 367 = 6973 

∴ n = 38; S_n = 6973

To find: 10^th common term between the APs

Common difference of 1^st series = 4

Common difference of 2^nd series = 5

LCM of common difference will give us a common difference of new series

⟹ 5 × 4

⟹ 20

The first term of new AP will be 11, so the 10^th = term of this series is

⟹ 11 + 20 × 9

⟹ 191

Assume the numbers in AP are a - d, a, a + d 

Given that,

The sum of three numbers is 12 

To find : the first three terms of AP 

So, 

3a = 12 

a = 4 

It is also given that the sum of their cube is 288 

(a - d)³ + a³ + (a + d)³ = 288

a³ - d³ - 3ad(a - d) + a³ + a³ + d³ + 3ad(a + d) = 288 

Substituting a = 4 we get,

64 - d³ - 12d(4 - d) + 64 + 64 + d³ + 12d(4 + d) = 288 

192 + 24d² = 288 

d = 2 or d = - 2 

Hence, 

The numbers are a - d, a, a + d which is 2, 4, 6 or 6, 4, 2

Let the three numbers are in AP = a, a + d, a + 2d

According to the question,

The sum of three terms = 12

⇒ a + (a + d) + (a + 2d) = 12

⇒ 3a + 3d = 12

⇒ a + d = 4

⇒ a = 4 – d …(i)

and the sum of their cubes = 408

⇒ a³ + (a + d)³ + (a + 2d)³ = 408

⇒ (4 – d)³ + (4)³ + ( 4 – d + 2d)³ = 408 [from(i)]

⇒ (4 – d)³ + (4)³ + ( 4 + d)³ = 408

⇒ 64 – d³ + 12d² – 48d + 64 + 64 + d³ + 12d² + 48d = 408

⇒ 192 + 24d² = 408

⇒ 24d² = 408 – 192

⇒ 24d² = 216

⇒ d² = 9

⇒ d = √9

⇒ d = ±3

Now, if d = 3, then a = 4 – 3 = 1

and if d = – 3, then a = 4 – ( – 3) = 4 + 3 = 7

So, the numbers are →

if a = 1 and d = 3

1, 4, 7

and if a = 7 and d = – 3

7, 4, 1

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = \(\frac{n}{2}\)[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

Given series. 2 + 4 + 6 + . . . + 200 which is an A.P

Where, a = 2 ,d = 4 – 2 = 2 and last term (a_n = l) = 200

We know that, a_n = a + (n – 1)d

So,

200 = 2 + (n – 1)2

200 = 2 + 2n – 2

n = \(\frac{200}{2}\) = 100

Now, for the sum of these 100 terms

S₁₀₀ = \(\frac{100}{2}\) [2 + 200]

= 50(202)

= 10100

Hence, the sum of terms of the given series is 10100.

a = 17, l = 350, d = 9 

Number of terms, n = \(\frac{1-a}{d} + 1\)

\(\frac{350-17}{9} + 1\)

\(\frac{333+9}{9} = \frac{342}{9} = 38\)

s_n= \(\frac{n}{2}\) (a + l)

\(\frac{38}{2}(350 + 17)\)

= 19 (367) = 6973

 A coin has two sides a head(H) and a tail(T).

∴ All possible outcomes are H, T.

Total number of outcomes = 2

Chances of getting a tail = 1

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of getting a tail P(T) \(\frac{Number\,of\,getting\,a\,tail}{Total\,number\,of\,outcomes}\) = \(\frac{1}{2}\)

Correct option is (A) 10 – n

Let a & d be first term & common difference of A.P.

We have \(a_{10}=0\;\&\;a_{42}=-32\)

\(\therefore a+9d=0\)         ______________(1)

& \(a+41d=-32\)   ______________(2)    \((\because a_n=a+(n-1)d)\)

Subtract equation (1) from (2), we get

\((a+41d)-(a+9d)=-32-0\)

\(\Rightarrow32d=-32\)

\(\Rightarrow d=\frac{-32}{32}=-1\)

\(\therefore a=-9d=-(-9)=9\)         (From (1) & d = -1)

\(\therefore a_n=a+(n-1)d\)

\(=9+(n-1)d\)

\(=9-n+1\)

\(=10-n\)

Hence, \(n^{th}\) term of A.P. is 10 - n.

Correct option is A) 10 – n

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = \(\frac{n}{2}\)[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

Given series. 3 + 11 + 19 + . . . + 803 which is an A.P

Where, a = 3 ,d = 11 – 3 = 8 and last term (a_n = l) = 803

We know that, a_n = a + (n – 1)d

So,

803 = 3 + (n – 1)8

803 = 3 + 8n – 8

n = \(\frac{808}{8}\) = 101

Now, for the sum of these 101 terms

S₁₀₁ = \(\frac{101}{2}\) [3 + 803]

= \(\frac{101(806)}{2}\)

= 101 x 403

= 40703

Hence, the sum of terms of the given series is 40703.

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = \(\frac{n}{2}\)[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

Given series. 1 + 3 + 5 + 7 + . . . + 199 which is an A.P

Where, a = 1 ,d = 3 – 1 = 2 and last term (a_n = l) = 199

We know that, a_n = a + (n – 1)d

So,

199 = 1 + (n – 1)2

199 = 1 + 2n – 2

n = \(\frac{200}{2}\) = 100

Now, for the sum of these 100 terms

S₁₀₀ = \(\frac{100}{2}\) [1 + 199]

= 50(200)

= 10000

Hence, the sum of terms of the given series is 10000.

Given, in an A.P

a₃ = 16 and a₇ = a₅ + 12

We know that a_n = a + (n – 1)d

⇒ a + 2d = 16…… (i)

And,

a + 6d = a + 4d + 12

2d = 12

⇒ d = 6

Using d in (i), we have

a + 2(6) = 16

a = 16 – 12 = 4

Hence, the A.P is 4, 10, 16, 22, …….

Correct option is A) -13.5

Let’s consider the first term as a and the common difference as d.

Given,

a₃ = 7 …. (1) and,

a₇ = 3a₃ + 2   …. (2)

So, using (1) in (2), we get,

a₇ = 3(7) + 2 = 21 + 2 = 23  …. (3)

Also, we know that

a_n = a +(n – 1)d

So, the 3th term (for n = 3),

a₃ = a + (3 – 1)d

⟹ 7 = a + 2d   (Using 1)

⟹ a = 7 – 2d     …. (4)

Similarly, for the 7th term (n = 7),

a₇ = a + (7 – 1) d 24 = a + 6d = 23  (Using 3)

a = 23 – 6d  …. (5)

Subtracting (4) from (5), we get,

a – a = (23 – 6d) – (7 – 2d)

⟹ 0 = 23 – 6d – 7 + 2d

⟹ 0 = 16 – 4d

⟹ 4d = 16

⟹ d = 4

Now, to find a, we substitute the value of d in  (4), a =7 – 2(4)

⟹ a = 7 – 8

a = -1

Hence, for the A.P. a = -1 and d = 4

For finding the sum, we know that

S_n = \(\frac{n}{2}\)[2a + (n − 1)d] and n = 20 (given)

S₂₀  = \(\frac{20}{2}\)[2(−1) + (20 − 1)(4)]

= (10)[-2 + (19)(4)]

= (10)[-2 + 76]

= (10)[74]

= 740

Hence, the sum of first 20 terms for the given A.P. is 740

\(\frac{2}{9}, \frac{2}{3}, 2,.....\)

Given sequence is a geometric progression.

whose first term a \(=\frac{2}{9}\)

Common ratio r \(=\frac{a_2}{a_1}\) \(=\frac{\frac{2}{3}}{\frac{2}{9}}=\frac{2}{3}\times \frac{9}{2}\) = 3

\(\therefore\) \(7^{th}\) term of G.P is \(a_7 = ar^{7-1}=ar^6\)

\(=\frac{2}{9}\times 3^6=2\times 3^4\)

\(=2\times 81\)

= 162.

Correct option is D) \(\cfrac{n(n+1)}2\) - \(\cfrac{x(1-x^n)}{1-x}\)

Correct option is (B) \(9 \sum n^2 - 3 \sum n - 2n\)

1, 4, 7, 10, ........ will form an A.P. whose \(n^{th}\) term is \(a_n=a+(n-1)d\)

\(=1+(n-1)3\)

\(=3n-3+1\)

\(=3n-2\)

Also 4, 7, 10, 13, ......... will form another A.P. whose \(n^{th}\) term is \(a_n=a+(n-1)d\)

\(=4+(n-1)3\)

\(=3n-3+n\)

\(=3n+1\)

Now, 1.4+4.7+7.10+10.13+.........+n terms

= 1.4+4.7+7.10+10.13+.........+(3n-2)(3n+1)

\(=\sum(3n-2)(3n+1)\)

\(=\sum(9n^2-6n+3n-2)\)

\(=\sum(9n^2-3n-2)\)

\(=9\sum n^2-3\sum n-2\sum1\)

\(=9\sum n^2-3\sum n-2n\)         \((\sum1=n)\)

Correct option is B) 9Σn² – 3Σn – 2n 

Given : A man saved 16500/- in ten years 

To find : His saving in the first year 

Let he saved Rs. x in the first year 

Since each year after the first he saved 100/- more then he did in the preceding year 

So, 

A.P will be x, 100 + x, 200 + x……………….. 

Where x is first term and 

common difference, 

d = 100 + x – x = 100 

We know, 

S_n is the sum of n terms of an A.P

Formula used :

s_n = \(\frac{n}{2}\){2a + (n-1)d}

Where a is first term, d is common difference and n is number of terms in an A.P. 

According to the question : 

S_n = 16500 and n = 10 

Therefore,

s₁₀ = \(\frac{10}{2}\){2x + (10-1)100}

⇒ 16500 = 5{2x + 9(100)} 

⇒ 16500 = 5(2x + 900) 

⇒ 16500 = 10x + 4500 

⇒ -10x = 4500 – 16500 

⇒ –10x = –12000

⇒ x = \(\frac{-12000}{-10}\)

⇒ x = 1200

Hence, his saving in first year is 1200.

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = \(\frac{n}{2}\)[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

Given series (-5) + (-8) + (-11) + . . . + (- 230) which is an A.P

Where, a = -5 ,d = -8 – (-5) = -3 and last term (a_n = l) = -230

We know that, a_n = a + (n – 1)d

So,

-230 = -5 + (n – 1)(-3)

-230 = -5 – 3n + 3

3n = -2 + 230

n = \(\frac{228}{3}\) = 76

Now, for the sum of these 76 terms

S₇₆ = \(\frac{76}{2}\) [-5 + (-230)]

= 38 x (-235)

= -8930

Hence, the sum of terms of the given series is -8930.

Given x, x + 2 and x + 6 are in G.P. but read it as x, x + 2 and x + 6. 

∴ r = \(\frac{t_2}{t_1}\) = \(\frac{t_3}{t_2}\)

⇒ \(\frac{x+2}{x}\)= \(\frac{x+6}{x+2}\)

⇒(x + 2)² = x(x + 6) 

⇒ x² + 4x + 4 = x² + 6x 

⇒ 4x – 6x = – 4 = -2x = -4 

∴ x = 2

Given G.P. = 2, -6, 18, -54, ……. 

a = 2, r = \(\frac{a_2}{a_1}\) = \(\frac{-6}{2}\) = -3 

a_n = a . r^n-1 = 2 × (-3)^n-1 

∴ r = -3; a_n = 2(-3)^n-1

Correct option is (A) 1/64

Given sequence is \(1 , \frac{-1}{2}, \frac{1}{4}, ........\) 

\(\therefore a_1=1,a_2=\frac{-1}2,a_3=\frac14\)

Now, \(\frac{a_2}{a_1}=\frac{\frac{-1}2}1=\frac{-1}2,\)

\(\frac{a_3}{a_2}=\cfrac{\frac14}{\frac{-1}2}\)

\(=\frac14\times-2=\frac{-1}2\)

\(\because\) \(\frac{a_3}{a_2}=\frac{a_2}{a_1}\)

\(\therefore\) \(1 , \frac{-1}{2}, \frac{1}{4}, ........\) will form a G.P. with common difference \(\frac{-1}2.\)

\(\therefore r=\frac{a_2}{a_1}=\frac{-1}2\)

\(\therefore7^{th}\) term of G.P. \(=a_7=ar^6\)

\(=1.(\frac{-1}2)^6=\frac1{2^6}\)

\(=\frac1{64}\)

Correct option is A) 1/64

Correct option is (D) 3

We have \(a_3=7\;\&\;a_7=15\)

\(\therefore a+2d=7\)    ________________(1)

& \(a+6d=15\)  ________________(2)    \((\because a_n=a+(n-1)d)\)

Multiply equation (1) by (3), we get

\(3a+6d=21\)   ________________(3)

Subtract equation (2) from (3), we get

\((3a+6d)-(a+6d)=21-15\)

\(\Rightarrow2a=6\)

\(\Rightarrow a=\frac62=3\)

Hence, first term of A.P. is 3.

Correct option is D) 3

Let A₁, A₂, A₃, A₄, A₅ be the 5 nos Between 8 And 26 

Then, 

8, A₁, A₂, A₃, A₄, A₅, 26 are in AP 

We know, 

A_n = a + (n - 1)d 

A₇ = 26 = 8 + (7 - 1)d 

d = 3 

Hence, 

A₁ = a + d 

= 8 + 3 

= 11 

A₂ = A₁ + d 

= 11 + 3 

= 14 

A₃ = A₂ + d 

= 14 + 3 

= 17 

A₄ = A₃ + d 

= 17 + 3 

= 20 

A₅ = A₄ + d 

= 20 + 3 

= 23

Sum of the first 7 terms of AP = 98 

\(\frac{7}{2}\)[2a + (7-1)d] = 98 

2a + 6d = 98 × \(\frac{2}{7}\)

2a + 6d = 28 

a + 3d = 14 ……………..(1)

Sum of the first 15 terms of AP = 390 

\(\frac{15}{2}\) [2a + (15 – 1)d] = 390 

2a + 14d = 390 × \(\frac{2}{15}\)

2a + 14d – 52 

a + 7d = 26 ……………(2) 

by solving (1) and (2) 

a = 5 and d = 3 

Sum of the first 10 terms 10 

= \(\frac{10}{2}\) [2a + (10 – 1)d] 

– 5[2(5) + 9(3)] 

= 5[10 + 27] 

= 5 × 37 = 185

Given G.P.: √3, 3, 3√3, …….. is 729 

a = √3, r = \(\frac{a_2}{a_1}\) = \(\frac{3}{\sqrt{3}}\) = √3 

now a_n = a . r^n-1 = 729 

⇒ (√3)(√3)^n-1 = 729 

⇒ (√3)ⁿ = 3⁶ = (√3)¹² 

⇒ n = 12 

So 12th term of GP √3, 3, 3√3, …….. is 729.

Correct option is (B) \(\frac{(3^{n+1}-2n-3)}{4}\)

\(S_n=1+4+13+40+......+\) upto n terms

\(=1+(1+3)+(1+3+9)+(1+3+9+27)+........+\) upto n terms

\(=3^0+(3^0+3^1)+(3^0+3^1+3^2)\) \(+(3^0+3^1+3^2+3^3)+..........\) \(+(3^0+3^1+3^2+........+3^{n-1})\)

\(=\sum(3^0+3^1+3^2+........+3^{n-1})\)

\(=\sum\frac{1(3^n-1)}{3-1}\)            \((\because\) Sum in G.P. \(=\frac{a(r^n-1)}{r-1})\)

\(=\frac12(\sum 3^n-\sum1)\)

\(=\frac12\left(\frac{3(3^n-1)}{3-1}-n\right)\)         \(\left(\because\sum 3^n=\frac{3(3^n-1)}{3-1}\right)\)

\(=\frac14(3^{n+1}-3-2n)\)

\(=\frac{3^{n+1}-2n-3}{4}\)

Correct option is B) \(\cfrac{(3^{n+1}-2n-3)}{4}\)

Here,

a₁ = √3

a₂ = 2√3

a₃ = 3√3

a₄ = 4√3

a₂ – a₁ = 2√3 – √3 = √3

a₃ – a₂ = 3√3 – 2√3= √3

a₄ – a₃ = 4√3 – 3√3= √3

Since, difference of successive terms are equal,

Hence, √3 , 2√3, 3√3,… is an AP with common difference √3.

Therefore, the next three term will be,

4√3 + √3, 4√3 + 2√3, 4√3 + 3√3

5√3, 6√3, 7√3

A.P here is 3y – 1, 3y + 5 and 5y + 1. 

So d= common difference = a₂– a₁= a₃–a₂ 

3y + 5 – (3y–1) 

= (5y + 1) – (3y + 5) 6 

= 2y – 4 10 / 2 = y

y = 5

Given A.P: -5 + (-8) + (-11) + … + (-230) 

Here first term, a = -5; 

d = a₂ – a₁ = (-8) – (-5) = -8 + 5 = -3 and 

the last term l = a_n = 10 

But, a_n = a + (n – 1) d 

∴ (-230) = -5 + (n – 1) (-3) 

⇒ -230 + 5 = -3n + 3 

⇒ -3n + 3 = -225

⇒ -3n = -225 – 3 

⇒ 3n = 228 

⇒ n = 238/3 = 76 

∴ n = 76 

Now, S_n = \(\frac{n}{2}\)(a + l) 

where a = -5; l = -230 

S₇₆ = \(\frac{76}{2}\)((-5) + (-230)) 

= 38 × (-235) 

= -8930

In the given AP, first term, a = 72 and 

common difference, d = (68 - 72) = -4.

Let its T_n = 0

\(\Rightarrow\) a + (n - 1) d = 0

\(\Rightarrow\) 72 + (n - 1) x (-4) = 0

\(\Rightarrow\) 76 - 4n = 0

\(\Rightarrow\) 4n = 76

\(\Rightarrow\) n = 19

Hence, the 19^th term of the given AP is 0.

Correct answer is (b) 73

T₂ = a + d = 13    .......(i)

T₅ = a + 4d = 25  .......(ii)

On subtracting (i) from (ii), we get:

\(\Rightarrow\) 3d = 12

\(\Rightarrow\) d = 4

On putting the value of d in (i), we get:

\(\Rightarrow\) a + 4 = 13

\(\Rightarrow\) a = 9

Now, T₁₇ = a + 16d = 9 + 16 x 4 = 73

Hence, the 17^th term is 73.

Correct option is (D) 1/3

Let \(2,\frac{2}{3},\frac{2}{9},\frac{2}{27},.....\) be given sequence.

\(a_1=2,a_2=\frac23,a_3=\frac29,a_4=\frac2{27}\)

Now, \(\frac{a_2}{a_1}=\frac{\frac23}{2}=\frac{2}{6}=\frac{1}{3},\)

\(\frac{a_3}{a_2}=\cfrac{\frac{2}{9}}{\frac{2}{3}}=\frac{2}{9}\times\frac{3}{2}=\frac{1}{3},\)

\(\frac{a_4}{a_3}=\cfrac{\frac{2}{27}}{\frac{2}{9}}=\frac{2}{27}\times\frac{9}{2}=\frac{1}{3}\)

\(\because\frac{a_4}{a_3}=\frac{a_3}{a_2}=\frac{a_2}{a_1}=\frac13\)

\(\therefore\) Sequence \(2,\frac{2}{3},\frac{2}{9},\frac{2}{27},.....\) will form a G.P. whose common ratio is \(r=\frac{a_2}{a_1}=\frac13.\)

Correct option is D) 1/3

To find: Five numbers between 11 and 29, which are in A.P.

Given: (i) The numbers are 11 and 29

Formula used: (i) A_n = a + (n-1)d

Let the five numbers be A₁, A₂, A₃, A₄ and A₅

According to question 11, A₁, A₂, A₃, A₄, A₅ and 29 are in A.P.

We can see that the number of terms in this series is 7

For the above series:-

a = 11, n = 7

A₇ = 29

Using formula, A_n = a + (n-1)d

⇒ A₇ =11 + (7-1)d = 29

⇒ 6d = 29 – 11

⇒ 6d = 18

⇒ d = 3

We can see from the definition that A₁, A₂, A₃, A₄ and A₅ are five arithmetic mean between 11 and 29, where d = 3, a = 11

Therefore, Using formula of arithmetic mean i.e. A_n = a + nd

A₁ = a + nd = 11 + 3 = 14

A₂ = a + nd = 11 + (2)3 = 17

A₃ = a + nd = 11 + (3)3 = 20

A₄ = a + nd = 11 + (4)3 = 23

A₅ = a + nd = 11 + (5)3 = 26

(Ans) 14, 17, 20, 23 and 26 are the required numbers.

Yes, we can write the above said A.P. 

In the given A.P. (d) = 6 

Seventh term a₇ = 36 

a₇ = a + 6d = 36 

⇒ a + 6 (6) = 36 

∴ a = 36 – 36 = 0 

∴ In the given progression a = 0 and d = 6 then it is 0, 6, 12, 18, …………

 We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = \(\frac{n}{2}\)[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

Given series, 34 + 32 + 30 + . . . + 10 which is an A.P

Where, a = 34 ,d = 32 – 34 = -2 and last term (a_n = l) = 10

We know that, a_n = a + (n – 1)d

So,

10 = 34 + (n – 1)(-2)

10 = 34 – 2n + 2

n = \(\frac{(36 \,–\, 10)}{2}\) = 13

Now, for the sum of these 13 terms

S₁₃ = \(\frac{13}{2}\) [34 + 10]

= \(\frac{13(44)}{2}\)

= 13 x 22

= 286

Hence, the sum of terms of the given series is 286.

Correct option is D) \(\cfrac{n}{(n+1)}\)

Given A.P : 7 + 10\(\frac{1}{2}\) + 14 + …. + 84 

a = 7; d = a₂ – a₁ = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\) and the last term l = a_n = 84 

But, a_n = a + (n – 1) d

∴ 84 = 7 + (n – 1) 3\(\frac{1}{2}\) 

⇒ 84 – 7 = (n – 1) × \(\frac{7}{2}\) 

⇒ n – 1 = 77 × \(\frac{2}{7}\) = 22 

⇒ n = 22 + 1 = 23 

Now, S_n = \(\frac{n}{2}\) (a + l) where a = 7; l = 84 

S₂₃ = \(\frac{23}{2}\) (7 + 84) 

= \(\frac{23}{2}\) × 91 

= \(\frac{2093}{2}\)

= 1046\(\frac{1}{2}\)

Given A.P: 34 + 32 + 30 + … + 10 

a = 34; d = a₂ – a₁ = 32 – 34 = -2 and the last term l = a_n = 10 

But, a_n = a + (n – 1) d 

∴ 10 = 34 + (n – 1) (-2) 

⇒ 10 – 34 = -2n + 2 

⇒ -2n = -24 – 2 

⇒ n = -26/-2 = 13

∴ n = 13 

Also, S_n = \(\frac{n}{2}\) (a + l) 

where a = 34; l = 10 

S₁₃ = \(\frac{13}{2}\)(34 + 10) 

= \(\frac{13}{2}\) × 44 

= 13 × 22 

= 286