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Correct option is (D) 9

Given G.P. is \(3, 3 \sqrt3,9,.......\)

\(\therefore a_1=3,a_2=3\sqrt3\)

\(\therefore\) Common ratio is r \(=\frac{a_2}{a_1}\)

\(=\frac{3\sqrt3}3=\sqrt3\)

Let \(a_n=243\)

\(\therefore ar^{n-1}=243\)       \((\because a_n=ar^{n-1})\)

\(\Rightarrow3(\sqrt3)^{n-1}=243\)

\(\Rightarrow3\frac{n-1}2=\frac{243}3\)

\(=81=3^4\)

\(\Rightarrow\frac{n-1}2=4\)

\(\Rightarrow n-1=8\)

\(\Rightarrow n=8+1=9\)

Hence, \(9^{th}\) term of given G.P. equals to 243.

Correct option is D) 9

Correct option is (A) 16

In G.P., \(a_2=24\;\&\;a_5=81\)

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

\(\therefore ar=24\)        ______________(1)  \((\because a_2=ar)\)

and \(ar^4=81\)   ______________(2)  \((\because a_5=ar^4)\)

Divide equation (2) by (1), we get

\(\frac{ar^4}{ar}=\frac{81}{24}\)

\(\Rightarrow r^3=\frac{27}8\)

\(=\frac{3^3}{2^3}=(\frac32)^3\)

\(\Rightarrow\) \(r=\frac32\)

Put \(r=\frac32\) in equation (1), we get

\(\frac32a=24\)

\(\Rightarrow a=24\times\frac23=16\)

Hence, first term of G.P. is 16.

Correct option is A) 16

Correct option is (B) b/c

a, b, c are in G.P.

\(\therefore b=ar\)

and \(c=ar^2\)

\(\Rightarrow r=\frac ba\)

and \(\frac cb=\frac{ar^2}{ar}=r=\frac ba\)

\(\therefore\) \(\frac{a}{b}\) = \(\frac{b}{c}\)

Correct option is B) b/c

To Find: we need to find n when a_n = 0

Given: The series is 64, 60, 56, 52, 48, … and a_n = 0

a₁ = 64, a₂ = 60 and d = 60 – 64 = –4

(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 = 0 = a₁ + (n –1)(– 4)

0 – 64 = (n –1)(– 4) [subtract 64 on both sides]

– 64 = (n –1)(– 4)

64 = (n –1)4 [Divide both side by '–']

16 = (n –1) [Divide both side by 4]

n = 17^th [add 1 on both sides]

The 17^th term of this AP is equal to 0.

Correct option is (D) -13.5

Given that x, 2x+2, 3x+3 are in G.P.

\(\therefore(2x+2)^2=x(3x+3)\)

\(\Rightarrow4x^2+8x+4=3x^2+3x\)

\(\Rightarrow x^2+5x+4=0\)

\(\Rightarrow(x+1)(x+4)=0\)

\(\Rightarrow x+1=0\;or\;x+4=0\)

\(\Rightarrow x=-1\;or\;x=-4\)

Case I :-

If x = -1, then

\(2x+2=-2+2=0\)

\(3x+3=-3+3=0\)

\(\because-1,0,0\) are not in G.P.

but x, 2x+1, 3x+3 are not in G.P. which is contradiction.

Case II :-

If x = -4

Then \(2x+2=-8+2=-6\)

& \(3x+3=-12+3=-9\)

\(\because-4,-6,-9\) are in G.P. which is required.

\(\therefore r=\frac{-6}{-4}=\frac32\)

Next term \(=-9r=-9\times\frac32\)

\(=\frac{-27}2=-13.5\)

Hence, \(4^{th}\) term is -13.5

Correct option is D) -13.5

Correct option is (C) tan² α

We have common difference = d = 1

& \(6^{th}\) term \(=a_6=\frac{4\,cos^2\alpha+1}{cos^2\alpha}\)

Let a be the first term of A.P.

Then a+5d \(=\frac{4\,cos^2\alpha+1}{cos^2\alpha}\)    \((\because a_6=a+5d)\)

\(\Rightarrow\) \(a=\frac{4\,cos^2\alpha+1}{cos^2\alpha}-5d\)

\(=\frac{4\,cos^2\alpha+1}{cos^2\alpha}-5\)                \((\because d=1)\)

\(=\frac{4\,cos^2\alpha+1-5\,cos^2\alpha}{cos^2\alpha}\)

\(=\frac{1-cos^2\alpha}{cos^2\alpha}=\frac{sin^2\alpha}{cos^2\alpha}=tan^2\alpha\)     \((\because1-cos^2\alpha=sin^2\alpha)\)

Hence, first term of A.P. is \(tan^2 \alpha.\)

Correct option is C) tan² α 

Correct option is (A) 65

Let 10 arithmetic means between 3 & 10 are a+d, a+2d, a+3d, ........, a+10d.

\(\therefore\) 3, a+d, a+2d, a+3d, ........, a+10d, 10 will form an A.P.

\(\therefore a=3\;\&\;a+11d=10\)

\(\Rightarrow 11d=10-a\)

\(=10-3=7\)

\(\Rightarrow d=\frac7{11}\)

Now, sum of 10 arithmetic means

= (a+d) + (a+2d) + (a+3d) + ........ + (a+10d)

\(=10a+d\,(1+2+3+.......+10)\)

\(=10a+\frac{10\times11}2\,d\)          \((\because1+2+3+........+n=\frac{n(n+1)}2)\)

\(=10a+55d\)

\(=10\times3+55\times\frac7{11}\)          \((\because a=3\;\&\;d=\frac7{11})\)

\(=30+35=65\)

Hence, sum of 10 arithmetic means between 3 & 10 is 65.

Correct option is A) 65

Correct option is (D) b-a/n+1

Let a+d, a+2d, .........., a+nd are n arithmetic means inserted between a and b.

\(\therefore\) a, a+d, a+2d, .........., a+nd, b will form an A.P.

\(\therefore\) \(a+(n+1)d=b\)

\(\Rightarrow(n+1)d=b-a\)

\(\Rightarrow\) \(d=\frac{b-a}{n+1}\)

\(\therefore\) Common difference is \(d=\frac{b-a}{n+1}.\)

Correct option is D) b - a/n + 1

Correct option is (A) 81

Let \(ar,ar^2,ar^3,ar^4,ar^5\) are 5 geometric means between \(\frac{1}{3}\) and 243.

\(\therefore\) \(\frac13,ar,ar^2,ar^3,ar^4,ar^5,243\) will form a G.P.

\(\therefore a=\frac13\;\&\;ar^6=243\)

\(\Rightarrow r^6=\frac{243}a=\frac{243}{\frac13}\)

\(=243\times3=729\)

\(\Rightarrow r^6=3^6\)

\(\Rightarrow r=3\)

\(\therefore\) \(ar=\frac13\times3=1,\)

\(ar^2=\frac13\times9=3,\)

\(ar^3=\frac13\times27=9,\)

\(ar^4=\frac13\times81=27,\)

\(ar^5=\frac13\times243=81\)

Hence, 1, 3, 9, 27 and 81 are 5 geometric means.

Hence, one of the 5 geometric mean is 81.

Correct option is A) 81

Correct option is (C) 12

Geometric mean of 6 and 24 \(=(ab)^\frac12\)

\(=(6\times24)^\frac12\)

\(=(2\times3\times2\times4\times3)^\frac12\)

\(=(2^2\times2^2\times3^2)^\frac12\)

\(=(12^2)^\frac12\) = 12

Hence, geometric mean of 6 and 24 is 12.

Correct option is C) 12

Correct option is (D) \( (-1)^n . 3.2^{n-1}\)

Given sequence is -3, 6, -12, 24, -48, .........

\(\Rightarrow-1\times3,(-1)^2\times3\times2,(-1)^3\times3\times2^2,\) \((-1)^4\times3\times2^3,(-1)^5\times3\times2^4,.......\)

\(\therefore n^{th}\) term of given sequence is \( (-1)^n\times3\times2^{n-1}\)

Correct option is D) (-1)ⁿ . 3.2^{n - 1}

Correct option is (C) 5n – 1

Let \(a_1=4,a_2=9,a_3=14\)

\(\because\) \(a_2-a_1\) \(=9-4=5\)

and \(a_3-a_2\) \(=14-9=5\)

\(\because\) \(a_3-a_2\) \(=a_2-a_1\)

\(\therefore\) 4, 9, 14, ......... will form an A.P. whose first term is \(a=a_1=4\) & common difference is \(d=a_2-a_1=9-4=5\)

\(\therefore\) \(n^{th}\) term of A.P. is \(a_n=a+(n-1)d\)

\(=4+(n-1)5\)

\(=4+5n-5\)

\(=5n-1\)

Hence, \(n^{th}\) term of 4+9+14+..….. is \(5n – 1.\)

Correct option is C) 5n – 1

let the three AM be x₁,x₂,x₃.

So new AP will be

6,x₁,x₂,x₃, - 6

Also - 6 = 6 + 4d

d = - 3

x₁ = 3

x₂ = 0

x₃ = - 3

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

And we know that, sum of first n terms is:

S_n = \(\frac{n}{2}\)(2a + (n − 1)d)

Also, nth term is given by a_n = a + (n – 1)d

From the question, we have

S_q = 162 and a₆ : a₁₃ = 1 : 2

So,

2a₆ = a₁₃

⟹ 2 [a + (6 – 1d)] = a + (13 – 1)d

⟹ 2a + 10d = a + 12d

⟹ a = 2d  …. (1)

And, S₉ = 162

⟹ S₉ = \(\frac{9}{2}\)(2a + (9 − 1)d)

⟹ 162 = \(\frac{9}{2}\)(2a + 8d)

⟹ 162 × 2 = 9[4d + 8d]  [from (1)]

⟹ 324 = 9 × 12d

⟹ d = 3

⟹ a = 2(3) [from (1)]

⟹ a = 6

Hence, the first term of the A.P. is 6

For the 15^th term, a₁₅ = a + 14d = 6 + 14 × 3 = 6 + 42

Therefore, a₁₅ = 48

Correct option is (C) 32,766

Let a & r be the first term & common difference of G.P.

\(\therefore\) r = common difference = 2

and \(a^5=32\)

\(\Rightarrow ar^4=32\)

\(\Rightarrow a=\frac{32}{r^4}=\frac{32}{2^4}\)

\(=\frac{32}{16}=2\)

\(\therefore\) Sum of 14 terms is \(S_{14}=\frac{a(r^{14}-1)}{r-1}\)

\(=\frac{2(2^{14}-1)}{2-1}\)

\(=2(16384-1)\)

\(=2\times16383=32766\)

Hence, sum of 14 terms in G.P. is 32766.

Correct option is C) 32,766

Correct option is (D) 7350

First & last even numbers between 100 & 200 are 102 & 198 respectively.

\(\therefore a_1=a=10^2\;\&\;d=2,a_n=198\)

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

\(\Rightarrow198=102+(n-1)2\)

\(\Rightarrow2(n-1)=198-102=96\)

\(\Rightarrow n-1=\frac{96}2=48\)

\(\Rightarrow n=48+1=49\)

\(\therefore\) Sum of even numbers between 100 & 200 is

\(S_n=\frac n2[a+a_n]\)

\(=\frac{49}2(102+198)\)

\(=\frac{49}2\times300\)

\(=49\times150=7350\)

Correct option is D) 7350

Let the first term, common difference of an AP are a and d respectively.

a = 7

d = a₂ - a₁ = 10 - 7 = 3

Let the nth term of this AP is 55

Then

a_n = 55

a + (n - 1)d = 55

7 + (n - 1)3 = 55

3(n - 1) = 48

n - 1 = 16

n = 17

So the 55 is a term of given AP and 55 is the 17^th term of given AP.

Given: 9^th term is 0

To prove: 29^th term is double the 19^th term

a + 8d = 0

a = - 8d

29^th term is

a + 28d

⟹ 20d

19^th term is

a + 18d

⟹ 10d

Hence proved 29^th term is double the 19^th term

(C) 25

Explanation:

Given, the common difference of AP i.e., d = 5

Now,

As we know, nth term of an AP is

a_{n =} a + (n – 1)d

where a = first term

a_n is nth term

d is the common difference

a₁₈ -a_{13 =} a + 17d – (a + 12d)

= 5d

= 5(5)

= 25

The correct option is (b) 2.

Explanation:

T₈ = a + 7d = 17 ...(i) 

T₁₄ = a + 13d = 29 ...(ii) 

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

⇒ 6d = 12 

⇒ d = 12/6

⇒ d = 2

Hence, the common difference is 2.

Given: a = 7; 

a₁₃ = a + 12d = 35 

⇒ 7 + 12d = 35 

⇒ 12d = 35 – 7 

⇒ n = \(\frac{28}{12}\) = \(\frac{7}{3}\)

Now, S_n = \(\frac{n}{2}\)(a + l) 

S₁₃ = \(\frac{13}{2}\)(7 + 35) 

= \(\frac{13}{2}\) × 42 

= 13 × 21 

= 273

Given, a_n = -4n + 15

Now putting n = 1, 2, 3, 4 we get,

a₁ = -4(1) + 15 = -4 + 15 = 11

a₂ = -4(2) + 15 = -8 + 15 = 7

a₃ = -4(3) + 15 = -12 + 15 = 3

a₄ = -4(4) + 15 = -16 + 15 = -1

We can see that,

a₂ – a₁ = 7 – (11) = -4

a₃ – a₂ = 3 – 7 = -4

a₄ – a₃ = -1 – 3 = -4

Since the difference between the terms is common, we can conclude that the given sequence defined by a_n = -4n + 15 is an A.P with common difference of -4.

Hence, the 15^th term will be

a₁₅ = -4(15) + 15 = -60 + 15 = -45

Correct option is (D) 225

We have d = 2 & \(a_8=15\)

\(\therefore a+7d=15\)        \((\because a_8=a+(8-1)d=a+7d)\)

\(\Rightarrow a+7\times2=15\)

\(\Rightarrow a=15-14=1\)

\(\therefore\) Sum of 15 terms is \(S_{15}=\frac{15}2[2a+(15-1)d]\)

\(=\frac{15}2[2\times1+14\times2]\)         \((\because a=1\;\&\;d=2)\)

\(=\frac{15}2(2+28)\)

\(=\frac{15}2\times30\)

\(=15\times15=225\)

Correct option is D) 225

Given, a_n = 5n – 7

Now putting n = 1, 2, 3, 4 we get,

a₁ = 5(1) – 7 = 5 – 7 = -2

a₂ = 5(2) – 7 = 10 – 7 = 3

a₃ = 5(3) – 7 = 15 – 7 = 8

a₄ = 5(4) – 7 = 20 – 7 = 13

We can see that,

a₂ – a₁ = 3 – (-2) = 5

a₃ – a₂ = 8 – (3) = 5

a₄ – a₃ = 13 – (8) = 5

Since the difference between the terms is common, we can conclude that the given sequence defined by a_n = 5n – 7 is an A.P with common difference 5.

Correct option is (A) 32,766

Let first term of G.P. be a.

Given that common ratio is r = 2.

\(\therefore5^{th}\) term of G.P. is \(a_5=ar^4\)       \((\because a_n=ar^{n-1})\)

\(\Rightarrow ar^4=32\)

\(\Rightarrow a.2^4=32\)

\(\Rightarrow a=\frac{32}{2^4}=\frac{32}{16}=2\)

Sum of first 14 terms is \(S_{14}=\frac{a(r^{14}-1)}{r-1}\)    \(\left(\because S_{n}=\frac{a(r^{n}-1)}{r-1}\right)\)

\(=\frac{2(2^{14}-1)}{2-1}\)                 \((\because a=2\;\&\;r=2)\)

= 2 (16384 - 1)           \((\because2^{14}=16384)\)

= 32766

Correct option is A) 32, 766

To find: number of terms in AP

Also

d = 16 – 13

d = 3

Also

43 = 13 + n × 3 – 3

So

n = 11

The first 2 digit number divisible by 7 is 14, and the last 2 digit number divisible by 7 is 98, so it forms AP with common difference 7

14,…,98

98 = 14 + (n - 1) × 7

n = 22

Correct answer is (b) 2

T₈ = a + 7d = 17     .......(i)

T₁₄ = a + 13d = 29    ........(ii)

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

\(\Rightarrow\) 6d = 12

\(\Rightarrow\) d = 12

Hence, the common difference is 2.

Correct option is (A) 8

Given that sum of n terms of an A.P. is \(4n^2+5n.\)

\(\therefore\) \(S_n=4n^2+5n\)

\(\therefore a_1=S_1=4.1^2+5.1\)

\(=4+5=9\)

\(\&\;S_2=4.2^2+5.2\)

\(=16+10=26\)

\(\Rightarrow a_1+a_2=26\)       \((\because S_2=a_1+a_2)\)

\(\Rightarrow a_2=26-a_1\)

\(=26-9=17\)

Now, \(d=a_2-a_1\)

\(=17-9=8\)

Hence, common difference of given A.P. is 8.

Correct option is A) 8

(i) Put n = 1

A₁ = 3 + 4(1) = 7

Put n = 2

A₂ = 3 + 4(2) = 11

Put n = 3

A₃ = 3 + 4(3) = 15

Common difference,

d₁ = a₂ – a₁ = 11 – 7 = 4

Common difference,

d₂ = a₃ – a₂ = 15 – 11 = 4

Since,

d₁ = d₂

Therefore, it’s an A.P. with sequence 7, 11, 15,… 

(ii) Put n = 1

A₁ = 5 + 2(1) = 7

Put n = 2

A₂ = 5 + 2(2) = 9

Put n = 3

A₃ = 5 + 2(3) = 11

Common difference,

d₁ = a₂ – a₁= 9 – 7 = 2

Common difference,

d₂ = a₃ – a₂= 11 – 9 = 2

Since,

d₁ = d₂

Therefore, it’s an A.P. with sequence 7, 9, 11,… 

(iii) Put n = 1

A₁ = 6 – 1 = 5

Put n = 2

A₂ = 6 – 2 = 4

Put n = 3

A₃ = 6 – 3 = 3

Common difference,

d₁ = a₂ – a₁= 4 – 5 = -1

Common difference,

d₂ = a₃ - a₂= 3 – 4 = -1

Since,

d₁ = d₂

Therefore, it’s an A.P. with sequence 5, 4 , 3,… 

(iv) Put n = 1

A₁ = 9 – 5(1) = 4

Put n = 2

A₂ = 9 – 5(2) = -1

Put n = 3

A₃ = 9 – 5(3) = -6

Common difference,

d₁ = a₂ – a₁ = -1 – 4 = -5

Common difference,

d₂ = a₃ – a₂ = - 6 – (-1) = - 5

Since,

d₁ = d₂

Therefore, it’s an A.P. with sequence 4, -1, -6,…

To find: 8^th term from the end

d = 9 - 7

d = 2

Also

201 = 7 + n × 2 – 2

n = 98

So 8^th term from end will be

7 + 90 × 2

⟹ 187

As n – 2, 4n – 1, 5n + 2 are in AP, 

so (4n – 1) – (n – 2) = (5n + 2) – (4n – 1) 

i.e, 3n + 1 = n + 3 

i.e, n = 1

Here, a = –11, d = –7 – (–11) = 4, a_n = 49 

We have an = a + (n – 1) d 

So, 49 = –11 + (n – 1) × 4 

i.e., 60 = (n – 1) × 4 

i.e., n = 16 

As n is an even number, there will be two middle terms which are 16/2th and (16/2 + 1)th, i.e., the 8th term and the 9th term.

a₈ = a + 7d = –11 + 7 × 4 = 17 

a₉ = a + 8d = –11 + 8 × 4 = 21 

So, the values of the two middle most terms are 17 and 21, respectively.

Given AP is AP 11, 8, 5, 2,…

Here a = 11, d = 8-11= -3

Let -150 be the nth term of the AP

a_n = a + (n – 1)d

-150 = 11 +(n-1)(-3)

or n = 164/3

Which is a fraction.

Therefore, -150 is not a term of the given AP.

AP is 10, 7, 4, …..…, (-62)

a = 10,

d = 7 – 10 = -3, and

l = -62

Now, a_n = l = a + (n – 1)d

-62 = 10 + (n – 1) x (-3)

-62 – 10 = -3(n- 1)

-72 = -3(n – 1)

Or n = 24 + 1 = 25

Middle term = (25 + 1)/2 th = 13th term

Find the 13^th term using formula, we get

a₁₃ = 10 + (13 – 1)(-3) 

= 10 – 36 

= -26

Here, a = 7 and d = (10 - 7) = 3, l = 184 and n = 8^th from the end.

Now, n^th term from the end = \([I-(n\,-1)d]\)

8^th term from the end = \([184-(8\,-1)\times3]\)

= \([184-(7\times3)]=(184\,-\,21)=163\)

Hence, the 8^th term from the end is 163.

Here, a = 7, d = 10 – 7 = 3 and l = 184

where l = a + (n – 1)d

Now, to find the 8^th term from the end, we will find the total number of terms in the AP.

So, 184 = 7 + (n – 1)(3)

⇒ 184 = 7 + 3n – 3

⇒ 184 = 4 + 3n

⇒ 180 = 3n

⇒ n = 60

So, there are 60 terms in the given AP.

Last term = 60^th

Second last term = 60 – 1 = 59^th

Third last term = 60 – 2 = 58^th

And so, on

So, the 8^th term from the end = 60 – 7 = 53^th term

So, a_n = a + (n – 1)d

⇒ a₅₃ = 7 + (53 – 1)(3)

⇒ a₅₃ = 7 + 52 × 3

⇒ a₅₃ = 7 + 156

⇒ a₅₃ = 163

Given: AP is 7, 10, 13,…, 184

a = 7, d = 10 – 7 = 3 and l = 184

n^th term from the end = l – (n – 1)d

Now,

8th term from the end be

184 – (8 – 1)3 

= 184 – 21

= 163

Given AP is 3, 7, 11, 15, …

a = 3, d = 7 – 3 = 4

Let 184 be the nth term of the AP

a_n = a + (n – 1)d

184 = 3 + (n – 1) x 4

184 – 3 = (n – 1) x 4

181 /4 = n – 1

n = 181 /4 + 1 = 185/4 (Which is in fraction)

Therefore, 184 is not a term of the given AP.

Correct option is (B) 2, 6, 18

Let required three numbers in G.P. are a, ar & \(ar^2.\)

\(\therefore\) Their sum \(=a+ar+ar^2\)

\(\therefore\) \(a+ar+ar^2=26\)              (Given)

\(\Rightarrow a(1+r+r^2)=26\)     ______________(1)

Also \(a.ar+ar.ar^2+a.ar^2=156\)     (Given)

\(a^2r(1+r+r^2)=156\)    ______________(2)

Divide equation (2) by (1), we get

\(\frac{a^2r(1+r+r^2)}{a(1+r+r^2)}=\frac{156}{26}\)

\(\Rightarrow ar=6\)                     ______________(3)

Then from (1), we have

\(a+ar+ar.r=26\)

\(\Rightarrow a+6+6r=26\)         \((\because ar=6)\)

\(\Rightarrow a+6r=26-6=20\)

\(\Rightarrow ar+6r^2=20r\)

\(\Rightarrow6+6r^2=20r\)          \((\because ar=6)\)

\(\Rightarrow6r^2-20r+6=0\)

\(\Rightarrow3r^2-10r+3=0\)

\(\Rightarrow3r^2-9r-r+3=0\)

\(\Rightarrow3r(r-3)-1(r-3)=0\)

\(\Rightarrow(r-3)(3r-1)=0\)

\(\Rightarrow r-3=0\;or\;3r-1=0\)

\(\Rightarrow r=3\;or\;r=\frac13\)

From (3), we have

\(a=\frac6r=\frac63=2\) or

\(a=\frac6r=\frac6{\frac13}=18\)

If r = 3 then a = 2

or if \(r=\frac13\) then a = 18

Case I :-

r = 3 & a = 2

\(\therefore ar=2\times3=6\)

\(ar^2=2\times9=18\)

Hence, required number are 2, 6 and 18.

Case II :-

If \(r=\frac13\) & a = 18

\(\therefore ar=\frac{18}3=6\)

& \(ar^2=\frac{18}9=2\)

Hence, required numbers in G.P. are 2, 6 & 18 or 18, 6, 2.

Alternate :-

Let \(a,ar,ar^2\) be required three numbers in G.P.

\(\therefore a+ar+ar^2=26\)

\(\Rightarrow a(1+r+r^2)=26\)         _______________(1)

And \(a.ar+ar.ar^2+ar^2.a=156\)

\(\Rightarrow a^2r(1+r+r^2)=156\)    _______________(2)

Divide equation (2) by (1), we get

\(\frac{a^2r(1+r+r^2)}{a(1+r+r^2)}=\frac{156}{26}=6\)

\(\Rightarrow ar=6\)            _______________(3)

Put ar = 6 in equation (1), we get

\(a+6+6r=26\)

\(\Rightarrow a+6r=26-6=20\)

\(\Rightarrow ar+6r^2=20r\)       (Multiply both sides by r)

\(\Rightarrow6+6r^2-20r=0\)

\(\Rightarrow3r^2-10r+3=0\)

\(\Rightarrow3r^2-9r-r+3=0\)

\(\Rightarrow3r(r-3)-1(r-3)=0\)

\(\Rightarrow(r-3)(3r-1)=0\)

\(\Rightarrow r-3=0\;or\;3r-1=0\)

\(\Rightarrow r=3\;or\;r=\frac13\)

Case I :-

r = 3 then \(a=\frac6r\)

\(=\frac63=2\)

\(\therefore ar=2\times3=6\)

\(ar^2=2\times9=18\)

Case II :-

\(r=\frac13\) then \(a=\frac6r=\frac6{\frac13}\)

\(=6\times3=18\)

\(\therefore ar=18\times\frac13=6\)

and \(ar^2=18\times\frac19=2\)

Hence, required numbers are 2, 6, 18.

Correct option is B) 2, 6, 18

Correct option is A) \(\cfrac{5n}9\) - \(\cfrac{5}{81}\) \((1-\cfrac{1}{10^n})\)

\(\sum_{i=1}^n \) a_i = 5n² - 3n

= 5n² + 5n - 8n

= 10\(\big(\frac{n^2+n}{2}\big)\)- 8n

=10 \(\sum_{i=1}^n \) i - 8 \(\sum_{i=1}^n \)1

= \(\sum_{i=1}^n \)(10 i -8)

i^th term of sequence = 10i - 8

100^th term of sequence = 1000-8 = 992

Sum of digits = 9 + 9 + 2 = 20

Given,

A.P is 84, 80, 76,… 

Here, 

a₁ = a = 84, 

a₂ = 88 

Common difference, 

d = a₂ – a₁ 

= 80 – 84 

= -4 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a₁ and d is common difference 

∴ a_n = 84 + (n – 1)-4 

⇒ a_n = 84 – 4n + 4 

⇒ a_n = 88 – 4n 

Now, 

To find which term of A.P is 0 

Put an = 0 

∴ 88 – 4n = 0 

⇒ -4n = -88

⇒ n = \(\frac{-88}{-4}\)

⇒ n = 22 

Hence, 

22^th term of given A.P is 0.

Given,

A.P is 3, 8, 13,… 

Here, 

a₁ = a = 3, 

a₂ = 8 

Common difference, 

d = a₂ – a₁ 

= 8 – 3 

= 5 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a1 and d is common difference 

∴ a_n = 3 + (n – 1)⁵ 

⇒ a_n = 3 + 5n – 5 

⇒ a_n = 5n – 2 

Now, 

To find which term of A.P is 248 

Put a_n = 248 

∴ 5n – 2 = 248 

⇒ 5n = 248 + 2 

⇒ 5n = 250 

⇒ n = \(\frac{250}{5}\)

⇒ n = 50 

Hence, 

50^th term of given A.P is 248.

Let common difference of an A.P is d and first term is a 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a1 and d is common difference.

Now, 

Take L.H.S.: 

a_m+n + a_m-n = a + (m + n – 1)d + a + (m - n – 1)d 

⇒ a_m+n + a_m-n = a + md + nd – d + a + md - nd – d 

⇒ a_m+n + a_m-n = 2a + 2md – 2d 

⇒ a_m+n + a_m-n = 2(a + md – d) 

⇒ a_m+n + a_m-n = 2[a + d(m – 1)] 

{∵ an = a + (n – 1)d} 

⇒ a_m+n + a_m-n = 2a_m 

Hence Proved.

Let the temperatures from Monday to Saturday in A.P. be 

a, a + d, a + 2d, a + 3d, a + 4d, a + 5d. 

According to the first condition, 

(a) + (a + 5d) = (a + d) + (a + 5d) + 5° 

∴ d = -5° 

According to the second condition, 

a + 2d = -30° 

∴ a + 2(-5°) = -30° 

∴ a – 10° = -30° 

∴ a = -30° + 10° = -20° 

∴ a + d = -20° – 5° = – 25° 

a + 3d = -20° + 3(- 5°) = -20° – 15° = -35° 

a + 4d = -20° + 4(-5°) = -20° – 20° = -40°

a + 5d = -20° + 5(-5°) = -20° – 25° = -45° 

∴ The temperatures on the other five days are -20°C, -25° C, -35° C, -40° C and -45° C.

a = 3, d = 15 – 3 = 12

Let last term be a_n

a_n = a + (n – 1) d

= 3 + (n – 1) 12

= 12n – 9 (i)

a₂₁ = a + 20d

= 3 + 20(12) = 243

Now, the term is 120 more than the 21^st term

a_n = 120 + a₂₁

= 120 + 243

= 363

Putting this value in (i), we get

363 = 12n – 9

12n = 363 + 9

n = 31

Hence, 31^st term of given A.P. is 120 more than its 21^st term

Correct option is (C) 7^th

Given G.P. is \(\frac{1}{3},\frac{1}{9},\frac{1}{27},........\)

\(\therefore a_1=\frac13,a_2=\frac19\)

\(\therefore\) Common ratio = r

\(=\frac{a_2}{a_1}=\cfrac{\frac19}{\frac13}\)

\(=\frac19\times3=\frac13\)

Let \(a_n\) = \(\frac{1}{2187}\)

\(\Rightarrow\) \(ar^{n-1}\) = \(\frac{1}{2187}\)     \((\because a_n=ar^{n-1}\) for GP)

\(\Rightarrow\) \(r^{n-1}\) = \(\frac{1}{2187a}\)

\(\Rightarrow\) \(r^{n-1}\) \(=\frac{3}{2187}=\frac1{729}\)   \((\because a=\frac13)\)

\(\Rightarrow(\frac13)^{n-1}=(\frac13)^6\)     \((\because r=\frac13)\)

\(\Rightarrow n-1=6\)          (By comparing)

\(\Rightarrow n=6+1=7\)

Hence, \(\frac{1}{2187}\) is \(7^{th}\) term of given G.P.

Correct option is C) 7^th

Correct option is (C) 137

\(\because a_1=-3\;\&\;a_2=4\)

\(\therefore d=a_2-a_1\)

= 4 - (-3)

= 4+3 = 7

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

\(\Rightarrow a_{21}=a_1+(21-1)d\)       \((\because n=21)\)

\(=-3+20\times7\)         \((a_1=-3\;\&\;d=7)\)

= -3+140 = 137

Correct option is C) 137

Correct option is (A) 8

We have \(a_{18} - a_{14} =32\)

\(\Rightarrow(a+17d)-(a+13d)=32\)       \((\because a_n=a+(n-1)d)\)

\(\Rightarrow4d=32\)

\(\Rightarrow d=\frac{32}4=8\)

Hence, the common difference of given A.P. is 8.

Correct option is A) 8