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Correct option is (A) 3, 5, 7, 9

Let a-3d, a-d, a+d, a+3d are required 4 numbers in A.P.

Given that sum of these four numbers is 24.

i.e., (a-3d) + (a-d) + (a+d) + (a+3d) = 24

\(\Rightarrow4a=24\)

\(\Rightarrow a=\frac{24}4=6\)

Also given that the sum of their squares is 164.

\(i.e.,(a-3d)^2+(a-d)^2\) \(+(a+d)^2+(a+3d)^2=164\)

\(\Rightarrow(a-3d)^2+(a+3d)^2\) \(+(a-d)^2+(a+d)^2=164\)

\(\Rightarrow2(a^2+(3d)^2)+2(a^2+d^2)=164\)     \((\because(a-b)^2+(a+b)^2=2(a^2+b^2))\)

\(\Rightarrow(a^2+9d^2)+(a^2+d^2)=\frac{164}2=82\)

\(\Rightarrow2a^2+10d^2=82\)

\(\Rightarrow10d^2=82-2a^2\)

\(\Rightarrow10d^2=82-2\times6^2\)       \((\because a=6)\)

\(=82-72=10\)

\(\Rightarrow d^2=\frac{10}{10}=1\)

\(\therefore d=1\)

Now, \(a-3d=6-3=3,\)

\(a-d=6-1=5,\)

\(a+d=6+1=7\)

and \(a+3d=6+3=9\)

Hence, required numbers are 3, 5, 7, 9.

Correct option is A) 3, 5, 7, 9

Correct option is (C) 10

24, 20, 16, …… is the arithmetic progression whose first term is \(a_1=a=24\;\&\)

common difference is \(d=a_2-a_1\)

\(=20-24=-4\)

Let sum of n terms is 60.

i.e., \(S_n=60\)

\(\Rightarrow\frac n2[2a+(n-1)d]=60\)

\(\Rightarrow n[2\times24+(n-1)\times-4]\)

\(=60\times2=120\)

\(\Rightarrow n(48-4n+4)=120\)

\(\Rightarrow n(52-4n)=120\)

\(\Rightarrow n(13-n)=\frac{120}4=30\)

\(\Rightarrow n^2-13n+30=0\)

\(\Rightarrow n^2-3n-10n+30=0\)

\(\Rightarrow n(n-3)-10(n-3)=0\)

\(\Rightarrow(n-3)(n-10)=0\)

\(\Rightarrow n-3=0\;or\;n-10=0\)

\(\Rightarrow n=3\;or\;n=10\)

Hence, sum of first 3 terms or sum of first 10 terms of given sequence is 60.

Correct option is C) 10

Given sequence whose a_n = 3ⁿ

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 in the above

a₁ = 3¹ = 3;

a₂ = 3² = 9;

a₃ = 3³ = 27;

a₄ = 3⁴ = 81;

a₅ = 3⁵ = 243.

∴ The required first five terms of the sequence whose n^th term, a_n = 3ⁿ are 3, 9, 27, 81, 243.

Given sequence whose a_n = (-1)ⁿ2ⁿ

To get first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above.

a₁ = (-1)¹.2¹ = (-1).2 = -2

a₂ = (-1)².2² = (-1).4 = 4

a₃ = (-1)³.2³ = (-1).8 = -8

a₄ = (-1)⁴.2⁴ = (-1).16 = 16

a₅ = (-1)⁵.2⁵ = (-1).32 = -32

∴ The first five terms of the sequence are – 2, 4, – 8, 16, – 32.

The first 15 multiples of 8 are 8, 16, 24, 32,…….

This is an AP in which a = 8, d = (16 - 8) = 8 and n = 15.

Thus, we have:

l = a + (n - 1)d

= 8 + (15 - 1)8

= 120

∴ Required sum = \(\frac{n}{2}(a+l)\)

= \(\frac{15}{2}[8+120]=15\times64=960\)

Hence, the required sum is 960.

Having common difference ‘d’, the AP 

1st set a, a+d, a+2d 

2nd set b, b + d, b + 2d 

100^th term of 1st set – 100^th term of 2nd set = 100 

∴ a + 99d – (b + 99d) = 100 

a + 99d – b – 99d = 100 a – b= 100 

Similarly. 

1000th term of 1st set = 1 + 999d 

1000 th term of 2nd set = b + 999d 

Their difference 

= a + 999d – (b + 999d) 

= a + 999d – b – 999d 

= a – b.

3, 15, 27, 39, ………… a_n = ?, n = ? 

a_n = a₅₄ + 132 

a = 3, d = 15 – 3 = 12 

a_n = a₅₄ + 132 

a_n = a + 53d + 132 

3 + 53(12) + 132 

= 3 + 636 + 132 

∴ a_n = 771 

a_n = a + (n – 1) d = 771 

= 3 + (n – 1)12 = 771 

3 + 12n — 12 = 771 

12n – 9 = 771 

12n = 771 + 9 

12n = 780 

∴ n = \(\frac{780}{12}\)

∴ n = 65. 

∴ 65^th term is 132 more than its 54^th term.

Given,

a = 5 and d = 3

We know that, n^th term a_n = a + (n – 1)d

So, for the given A.P. a_n = 5 + (n – 1)3 = 3n + 2

Also given, last term = 80

⇒ 3n + 2 = 80

3n = 78

n = 78/3 = 26

Therefore, there are 26 terms in the A.P.

Given A. P is 10.0, 10.5, 11.0, 11.5,

First term (a) = 10.0

Common difference (d) = Second term – first term

= 10.5 – 10.0 = 0.5

We know that, n^th term a_n = a + (n – 1)d

Then, 11^th term a₁₁ = 10.0 + (11 – 1)0.5

= 10.0 + 10 x 0.5

= 10.0 + 5

=15.0

∴ 11^th term of the A. P. is 15.0

Given,

a₆ = 19 and a₁₇ = 41

We know that, n^th term a_n = a + (n – 1)d

So,

a₆ = a + (6-1)d

⇒ a + 5d = 19 …… (i)

Similarity,

a₁₇ = a + (17 – 1)d

⇒ a + 16d = 41 …… (ii)

Solving (i) and (ii),

(ii) – (i) ⇒

a + 16d – (a + 5d) = 41 – 19

11d = 22

⇒ d = 2

Using d in (i), we get

a + 5(2) = 19

a = 19 – 10 = 9

Now, the 40^th term is given by a₄₀ = 9 + (40 – 1)2 = 9 + 78 = 87

Therefore the 40^th term is 87.

Given,

a₉ = 0

We know that, n^th term a_n = a + (n – 1)d

So, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ……(i)

Now,

29^th term is given by a₂₉ = a + (29 – 1)d

⇒ a₂₉ = a + 28d

And, a₂₉ = (a + 8d) + 20d [using (i)]

⇒ a₂₉ = 20d ….. (ii)

Similarly, 19^th term is given by a₁₉ = a + (19 – 1)d

⇒ a₁₉ = a + 18d

And, a₁₉ = (a + 8d) + 10d [using (i)]

⇒ a₁₉ = 10d …..(iii)

On comparing (ii) and (iii), it’s clearly seen that

a₂₉ = 2(a₁₉)

Therefore, 29^th term is double the 19^th term.

The correct option is: (C) 32

Explanation:

Series after the insertion of terms between 4 and 39 is 4, a₁, a₂, a₃, a₄, 39.

Now, 39 = 4 + 5d => 35 = 5d => d =7

. ^.. a₄ = 4 + 4 x 7 = 32

a = 1, d = 3

S_n=n/2[2a+(n-1)d]

S₂₀=20/2[2(1)+(20-1)3]

Three no. ‘s in A.P. be a - d, a, a + d

So,  a - d + a + a + d = - 3

3a = - 3 => a = -1 & (a - d) a (a + d) = 8

a(a² - d²) = 8

(-1) (1 - d²) = 8

1 - d² = - 8 => d² = 9 

=> d = ± 3

If a = 8 & d = 3 numbers are -4, -1, 2. 

If a = 8 & d = - numbers are 2, -1, -4.

407 = 3 + (n - 1)4 n = 102

So, 20^th term from end => m = 20

a_102-(20-1) = a_102-19 = a₈₃ from the beginning.

a₈₃ = 3 + (83 + 1)4 = 331.

Solution:

the first 40 positive integers divisible by 6 are 6,12,18,....... upto 40 terms

the given series is in arthimetic progression with first term a=6 and common difference d=6

sum of n terms of an A.p is 

n/2×{2a+(n-1)d}

→required sum = 40/2×{2(6)+(39)6}

=20{12+234}

=20×246

=4920

Answer is (C) 27^th

Given : S_n = 3n² + 5n

S₁ = 3(1)² + 5(1) = 8

S₂ = 3(2)² + 5(2) = 22

S₃ = 3(3)² + 5(3) = 42

S₄ = 3(4)² + 5(4) = 68

∴ a₁ = S₁ = 8

a₂ = S₂ – S₁ 

⇒ 22 – 8 ⇒ 14

a₃ = S₂ – S₁ 

⇒ 22 – 8 ⇒ 14

a₄ = S₂ – S₁ 

⇒ 22 – 8 ⇒ 14

Thus A.P. will be 8, 14, 20, 26 …… 164

a = 8,

d = 14 – 8 = 6 and a_n = 164

∴ 164 = a + (n – 1)d

164 = 8 + (n – 1)

(n – 1) = 156/6 = 26

∴ n = 26 + 1 = 27

To Find: AP and its 30^th term (i.e. a₃₀ = ?)

Given: a₅ = 5 and a₁₃ = –3

Formula Used: a_n = a + (n - 1)d

(Where a = a₁ is first term, a₂ is second term, an is n^th term and d is common difference of given AP)

By using the above formula, we have

a₅ = 5 = a + (5 - 1)d, and a₁₃ = –3 = a + (13 - 1)d

a + 4d = 5 and a + 12d = –3

On solving above 2 equation, we and a + 12d = –3 get

a = 9 and d= (–1)

So a₃₀ = 9 + 29(–1) = –20

AP is (9,8,7,6,5,4……) and 30^th term = –20

To Find: –47 is a term of the AP or not.

Given: The series is 5, 2, –1, –4, –7, ….

a₁ = 5, a₂ = 2, and d = 2 – 5 = –3 (Let suppose an = –47)

NOTE: n is a natural number.

(Where a = a₁ is first term, a₂ is second term, a_n is n^th term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a_n = –47 = a + (n - 1)d

–47 = 5 + (n - 1)(–3)

–47– 5 = (n - 1)(–3) [subtract 5 on both sides]

52 = (n - 1)(3) [Divide both side by ‘–‘]

17.33 = (n - 1) [Divide both side by 3]

18.33 = n [add 1 on both sides]

As n is not come out to be a natural number, So –47 is not the term of this AP.

The given AP is 2, 7, 12, ..., 47.

Let us re-write the given AP in reverse order i.e. 47, 42, .., 12, 7, 2.

Now, the 5th term from the end of the given AP is equal to the 5th term from beginning of

the AP 47, 42,.... ,12, 7, 2.

Consider the AP 47, 42,..., 12, 7, 2.

Here, a = 47 and d = 42 - 47 = -5

5th term of this AP

= 47 + (5 - 1) x (-5)

= 47 - 20

= 27

Hence, the 5th term from the end of the given AP is 27.

S_n = 3n² – n

S₁ = 3(1)² – 1 = 3 – 1 = 2

S₂ = 3(2)² – 2 = 12 – 2 = 10

a₁ = 2

a₂ = 10 – 2 = 8

(i) a_n = a + (n – 1) d

= 2 + (n – 1) x 6

= 2 + 6n – 6

= 6n – 4

(ii) First term = 2

(iii) Common difference = 8 – 2 = 6

AP is a, 3a, 5a,…..

Here, a = a, d = 2a

Sum = S_n = n/2 [2a + (n-1)d]

= n/2[2a + 2an – 2a]

= an²

Sum of first m terms of an AP = 2m² + 3m (given)

S_m = 2m² + 3m

Sum of one term = S₁ = 2(1)² + 3 x 1 = 2 + 3 = 5 = first term

Sum of first two terms = S₂ = 2(2)² + 3 x 2 = 8 + 6=14

Sum of first three terms = S₃ = 2(3)² + 3 x 3 = 18 + 9 = 27

Now,

Second term = a₂ = S₂ – S₁ = 14 – 5 = 9

Since, 

x² + xy + y², 

z² + zx + x² and 

y² + yz + z² are in AP 

(z² + zx + x²) - (x² + xy + y²) = (y² + yz + z²) - (z² + zx + x²) 

Let d = common difference, 

Since,

x, y, z are in AP 

Y = x + d and 

x = x + 2d 

Therefore the above equation becomes, 

= (x + 2d)² + (x + 2d)x - x(x + d) - (x + d)² 

= (x + d)² + (x + d)(x + 2d) - (x + 2d)x – x² 

= x² + 4xd + 4d² + x² + 2xd – x² – xd – x² - 2xd – d² 

= x² + 2dx + d² + x² + 2dx + xd + 2d² – x² - 2dx – x²

3xd + 3d² = 3xd + 3d²

Hence, proved.

x² + xy + y², z² + zx + x² and y² + yz + z² are in consecutive terms of an A.P.

a³ + c³ + 6abc = 8b³ 

a³ + c³ - (2b)³ + 6abc = 0 

a³ + (-2b)³ + c³ + 3a(-2b)c = 0 

Since, 

If a + b + c = 0, 

a³ + b³ + c³ = 3abc 

(a - 2b + c)³ = 0 

a - 2b + c = 0 

Since,

a, b, c are in AP 

b - a = c - b 

= a + c - 2b 

= 0 

Hence proved

(a - c)² = 4 (a - b) (b - c) 

a² + c² - 2ac = 4(ab – ac – b² + bc) 

a² + 4c²b² + 2ac - 4ab - 4bc = 0 

(a + c - 2b)² = 0 

a + c - 2b = 0 

Since,

a, b, c are in AP 

b - a = c - b 

a + c - 2b = 0 

Hence, 

(a - c)² = 4 (a - b) (b - c)

a² + c² + 4ac = 2 (ab + bc + ca) 

a² + c² + 2ac - 2ab - 2bc = 0 

(a + c - b)² – b² = 0 

a + c - b = b 

a + c - 2b = 0 

Since,

a, b, c are in AP 

b - a = c - b 

a + c - 2b = 0 

Hence proved.

Given: S_n = (2n² + 3n)

To find: find common difference

Put n = 1 we get

S₁ = 5 OR we can write

a = 5 …equation 1

Similarly put n = 2 we get

S₂ = 14 OR we can write

2a + d = 14

Using equation 1 we get

d = 4

We are given S_n=(2n²+3n)                                        ------(1)

To find the common difference 'd', d=a₂-a₁                ------(2)

First, to find the second term of the A.P., a₂=S₂ - S₁   --------(3)

Now, To find S₁, put n=1 in equation 1, we get,

S₁=(2.1²+3.1)=2+3=5                                                -----(4)

• Here S₁=a₁

Similarly for S₂,put n=1 in equation 1, we get,

S₂=(2.2²+3.2)=8+6=14                                             ------(5)

• Here S₂=a₂+a₁ 

 Using equation (3),(4) and (5) we get,

a₂=14-5=9

Now we know a₂=9 and a₁=5 ∵ a₁=S₁

From equation (2),

d=9-5=4

Hence, the common difference for the A.P.  is 4.

Wrong question. It will be 7n + 5 instead of 7n – 5.

Given: Ratio of sum of n terms of 2 AP’s

To Prove: 6^th terms of both AP’S are equal

Let us consider 2 AP series AP₁ and AP₂.

Putting n = 1, 2, 3… we get AP₁ as 12,19,26… and AP₂ as 22,27,32….

So, a₁ = 12, d₁ = 7 and a₂ = 22, d₂ = 5

For AP₁

S₆ = 12 + (6 - 1)7 = 47

For AP₂

S₆ = 22 + (6 - 1)5 = 47

Therefore their 6^th terms are equal.

Hence proved.

(i) a²(b + c), b²(c + a), c²(a + b) are also in A.P. 

b²(c + a) - a²(b + c) = c²(a + b) - b²(c + a)

b²c + b²a – a²b – a²c = c²a + c²b – b²a – b²c 

c(b² - a²) + ab(b - a) = a(c² - b²) + bc(c - b) 

(b - a)(ab + bc + ca) = (c - b)(ab + bc + ca) 

b - a = c - b 

And since a, b, c are in AP, 

b - a = c - b 

Hence given terms are in AP.

(ii) b + c - a, c + a - b, a + b - c are in A.P. 

Therefore, 

(c + a - b) - (b + c - a) = (a + b - c) - (c + a - b) 

2a - 2b = 2b - 2c 

b - a = c - b 

And since a, b, c are in AP, 

b - a = c - b 

Hence given terms are in AP.

(iii) bc – a², ca – b², ab – c² are in A.P. 

(ca - b²) – (bc - a²) = (ab - c²) – (ca - b²) 

(a - b)(a + b + c) = (b - c)(a + b + c) 

a - b = b - c 

b - a = c - b 

And since a, b, c are in AP, 

b - a = c - b 

Hence given terms are in AP

Here, a = 7 , d = 11 – 7 = 4 and l = 139

where l = a + (n – 1)d

⇒ 139 = 7 + (n – 1)(4)

⇒ 139 = 7 + 4n – 4

⇒ 139 = 3 + 4n

⇒ 139 – 3 = 4n

⇒ 136 = 4n

n = 136/4 = 34

Hence, the number of terms in the given AP is 34

We have

a₇ = a + (7 – 1)d = a + 6d = 20 …(1)

and a₁₃ = a + (13 – 1)d = a + 12d = 32 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 12d = 20 – 32

⇒ – 6d = –12

⇒ d = 2

Putting the value of d in eq (1), we get

a + 6(2) = 20

⇒ a + 12 = 20

⇒ a = 8

Hence, the required AP is 8, 10, 12, 14,…

For the given AP the first term a is a + b, and common difference d is a difference of the second term and the first term, which is a - b - (a + b) = - 2b 

To find : the sum of given AP 

The formula for sum of AP is given by,

s = \(\frac{n}{2}\)(2a + (n-1)d

Substituting the values in the above formula,

s = \(\frac{22}{2}\)(2a + 2b + (21)(-2b))

s = 11(2a - 40b)

For the given AP the first term a is 3, and common difference d is a difference of the second term and first term, which is 9/2-3 = 3/2.

To find : the sum of given AP 

The formula for sum of AP is given by,

s = \(\frac{n}{2}\)(2a + (n - 1)d

Substituting the values in the above formula,

s = \(\frac{25}{2}\)(6 + (24)(\(\frac{3}{2}\)))

s = 25 × 21 

s = 525

In the given AP, a = 41 and d = (38 - 41) = -3

Suppose that there are n terms in the given AP.

Then T_n = 8

\(\Rightarrow\) a + (n - 1) d = 8

\(\Rightarrow\) 41 + (n - 1) x (-3) = 8

\(\Rightarrow\) 3n = 36

\(\Rightarrow\) n = 12

Hence, there are 12 terms in the given AP.

Let the required numbers be (a - d), a and (a + d).

Then (a - d) + a + (a + d) = 3

\(\Rightarrow\) 3a = 3

\(\Rightarrow\) a = 1

Also, (a - d).a.(a + d) = -35

\(\Rightarrow\) a(a² - d²) = -35

\(\Rightarrow\) 1.(1 - d²) = -35

\(\Rightarrow\) d² = 36

\(\Rightarrow\) d = \(\pm6\)

Thus, a = 1 and d = \(\pm6\)

Hence, the required numbers are (-5,1 and 7) or (7,1 and -5).

Given: AP is 6, 10, 14, 18,…, 174

Here, first term = a = 6

Common difference = d= 10 – 6 = 4

To find: the number of terms (n)

Last term = a + (n – 1)d

174 = 6 + (n – 1) 4

174 – 6 = (n – 1) 4

n – 1 = 168 /4 = 42

n = 42 + 1 = 43

There are 43 terms.

For the given AP the first term a is 41, and common difference d is a difference of the second term and first term, which is 36 - 41 = - 5

To find : the sum of given AP 

The formula for sum of AP is given by,

s = \(\frac{n}{2}\)(2a + (n - 1)d)

Substituting the values in the above formula,

s = \(\frac{12}{2}\)(82 + (11)(-5))

s = 6 × 27 

s = 162

In the given AP, a = 6 d = (10 - 6) = 4

Suppose that there are n terms in the given AP.

Then, T_n = 174

\(\Rightarrow\) a + (n - 1) d = 174

\(\Rightarrow\) 6 + (n - 1) x 4 = 174

\(\Rightarrow\) 2 + 4n = 174

\(\Rightarrow\) 4n = 172

\(\Rightarrow\) n = 43

Hence, there are 43 terms in the given AP.

Given arithmetic series is  \(\frac{1}{5}\), \(\frac{3}{5}\), \(\frac{5}{5}\), \(\frac{7}{5}\), ………….

It is seen that, it’s of the form of  \(\frac{1}{5}\), \(\frac{3}{5}\), \(\frac{5}{5}\), \(\frac{7}{5}\), …………. a, a + d, a + 2d, a + 3d,

Thus, by comparing these two, we get

a = \(\frac{1}{5}\), a + d = \(\frac{3}{5}\), a + 2d = \(\frac{5}{5}\), a + 3d = \(\frac{7}{5}\)

First term (a) = \(\frac{1}{5}\)

By subtracting first term from second term, we get

d = (a + d)-(a)

d = \(\frac{3}{5}\) – \(\frac{1}{5}\)

d = \(\frac{2}{5}\)

⇒ common difference (d) = \(\frac{2}{5}\)

Let, three digit number divisible by 7 are

105, 112, 119,….994

Here, First term, a = 105

Common difference, d = 112 – 105 = 7

And last term, a_n = 994

a_n = a+ (n – 1) d

994 = 105 + (n – 1) 7

994 = 105 + 7n – 7

994 = 98 + 7n

7n = 896

n = 128

Hence, three digit number that are divisible by 7 are 128

a = 8, d = 6,

Last term = a_n

a_n = a + (n – 1) d

= 8 + (n – 1) 6

= 2 + 6n (i)

Let, 41^st term, a₄₁ = 8 + (41 – 1) 6

= 8 + 40 x 6 = 248

Now the term is 72 more than its 41^st term

a_n = 72 + a₄₁

= 72 + 248 = 320

Putting this value in (i), we get

320 = 2 + 6_n

6n = 318

n = 53

Hence, 53^rd term of the given A.P. is 72 more than its 41^st term

Given arithmetic series 0.3, 0.55, 0.80, 1.05, ……….

It is seen that, it’s of the form of a, a + d, a + 2d, a + 3d,

Thus, by comparing we get,

a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05

First term (a) = 0.3.

By subtracting first term from second term. We get

d = (a + d) – (a)

d = 0.55 – 0.3

d = 0.25

⇒ Common difference (d) = 0.25

General series is –1.1, – 3.1, – 5.1, –7.1, ……..

It is seen that, it’s of the form of a, a + d, a + 2d, a + 3d, ………..

Thus, by comparing these two, we get

a = –1.1, a + d = –3.1, a + 2d = –5.1, a + 3d = –71

First term (a) = –1.1

Common difference (d) = (a + d) – (a)

= -3.1 – ( – 1.1)

⇒ Common difference (d) = – 2

Let’s consider the three terms of the A.P. to be a – d, a, a + d

so, the sum of three terms = 21

⇒ a – d + a + a + d = 21

⇒ 3a = 21

⇒ a = 7

And, product of the first and 3rd = 2nd term + 6

⇒ (a – d) (a + d) = a + 6

a² – d² = a + 6

⇒ (7 )² – d² = 7 + 6

⇒ 49 – d² = 13

⇒ d² = 49 – 13 = 36

⇒ d² = (6)²

⇒ d = 6

Hence, the terms are 7 – 6, 7, 7 + 6 ⇒ 1, 7, 13

Correct answer is C. 21.

Let 3 consecutive terms A.P is a –d, a, a + d. and the sum is 51 

So, (a –d) + a + (a + d) = 51 

3a –d + d = 51 

3a = 51 

a = 17 

The product of first and third terms = 273 

So, (a –d) (a + d) = 273

 a² –d² = 273 

17² –d² = 273 

289 –d² = 273 

d² = 289 –273 

d² = 16 

d = 4 

Third term = a + d = 17 + 4 = 21

Correct option is B) (ii) only

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

\(1+8+27+64+......+n\) terms

\(=1^3+2^3+3^3+4^3+.........+n\) terms

\(=\left(\frac{n(n+1)}{2}\right)^2\)

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

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

Correct option is (B) \(\frac{n(4n^2+27n+59)}{6}\)

3.5 + 4.7 + 5.9 + ……. + n terms

= 3.5 + 4.7 + 5.9 + ……. + (n+2) (2n+3)

\((\because\) 3, 4, 5, ........ is an arithmetic progression whose \(n^{th}\) term is \(a_n=n+2\) & 5, 7, 9, ...... is an A.P. whose \(n^{th}\) term is \(a_n=(2n+3))\)

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

\(=\sum(2n^2+4n+3n+6)\)

\(=\sum(2n^2+7n+6)\)

\(=2\sum n^2+7\sum n+6\sum 1\)

\(=2\,\left(\frac{n(n+1)(2n+1)}6\right)+7\,\left(\frac{n(n+1)}2\right)+6n\)

\(\left(\because\sum n^2=\frac{n(n+1)(2n+1)}6,\sum n=\frac{n(n+1)}2\;\&\;\sum1=n\right)\)

\(=\frac n6(2(n+1)(2n+1)+21(n+1)+36)\)

\(=\frac n6(4n^2+6n+2+21n+21+36)\)

\(=\frac n6(4n^2+27n+59)\)

Correct option is B) \(\cfrac{n(4n^2+27n+59)}{6}\)

Correct option is (B) \(\sqrt3 ,\sqrt6 ,\sqrt{12} ,......\)

In a G.P. common ratio is \(r=\frac{a_2}{a_1}\)

(A) \(\sqrt2 ,\sqrt6 ,\sqrt{18} ,......\)

\(r=\frac{a_2}{a_1}=\frac{\sqrt6}{\sqrt2}\)

\(=\frac{\sqrt3.\sqrt2}{\sqrt2}=\sqrt3\)

(B) \(\sqrt3 ,\sqrt6 ,\sqrt{12} ,......\)

\(r=\frac{\sqrt6}{\sqrt3}\)

\(=\frac{\sqrt3.\sqrt2}{\sqrt3}=\sqrt2\)

(C) \(\sqrt5 ,\sqrt{15} ,\sqrt{45} ,......\)

\(r=\frac{a_2}{a_1}=\frac{\sqrt{15}}{\sqrt5}\)

\(=\frac{\sqrt3.\sqrt5}{\sqrt5}=\sqrt3\)

(D) \(\sqrt7 ,\sqrt{21} ,\sqrt{63} ,......\)

\(r=\frac{a_2}{a_1}=\frac{\sqrt{21}}{\sqrt7}\)

\(=\frac{\sqrt7.\sqrt3}{\sqrt7}=\sqrt3\)

Correct option is B) √3, √6, \(\sqrt{12}\),...........