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Let us say a be the first term and d be the common difference of an AP

a_n = a + (n – 1)d

a₇ = a + (7 – 1)d

= a + 6d = -4 …………(1)

And a₁₃ = a + 12d = -16 …..…..(2)

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

6d = -16 – (-4) = -12

From (1), a + 6d = -4

a + (-12) = -4

a = -4 + 12 = 8

a = 8, d = -2

AP will be 8, 6, 4, 2, 0, ……

(i) 3, 5, 7, 9, ..... 201

a = 3, d = 5 - 3 = 2, a_n = 201

a_n = a +(n -1)d

201 = 3 +2n – 2

N = 100

Now, we have to find 12^th term from the last that means,

100^th – 11 = 89^th term

Then,

a₈₉ = a + (89 – 1)d

= 3 + 88 x 2

= 179

Hence, the 12^th term from the end of the A.P. is 179.

(ii) 3, 8, 13, ....., 253

a = 3, d = 8 - 3 = 5

a_n = 253

a + (n - 1)d = 253

3 + (n - 1)5 = 253

n = 51

Now, we have to find 12th term from the last that means,

51^th -11 = 40^th term

Then,

a₄₀ = a + (n -1)d

= 3 + 39(5)

= 198

Hence, the 12^th term from the end of the A.P. is 198

(iii) 1, 4, 7, 10, ......., 88

a = 1, d = 4 -1 = 3

a_n = 88

a + (n -1)d = 88

1 + (n -1)3 = 88

n = 30

Now, we have to find 12^th term from the last that means,

30^th - 11 = 19^th term

a₁₉ = a + 18d

= 1 + 18(3)

= 55

Hence, the 12^th term from the end of the A.P. is 198.

a₆ = 19 = a + (n – 1) d

= 19 = a + 5d …(i)

A₁₇ = 41 = a + (n – 1) d

= 41 = a + 16d …(ii)

Subtracting (i) from (ii), we get

22 = 11d

d = 2

Now substituting the value of d in (i):

19 = a + 10

= a = 9

A₄₀ = a + (40 – 1) 2

= 9 + 78

= 87

10a₁₀ = 15a₁₅

10[a + (10 – 1)d] = 15[a + (15 – 1)d]

2a + 2(9)d = 3a + 3(14)d

- a = 42d – 18d

- a = 24d

a = - 24d

a₂₅ = a + 24d

= - 24d + 24d

= 0

a₉ = 0

a + (9 – 1)d =0

a + 8d = 0

a = - 8d …(i)

To Prove: a₂₉ = 2a₁₉

Proof: LHS= a₂₉ = a + 28d = - 8d + 28d = 20d

RHS= 2a₁₉ = 2[a + (18)d] = 2(- 8d + 18d) = 2(10d) = 20d

Since, LHS=RHS

Hence, proved

Put n = 1

a₁ = - 4(1)² + 15 = 11

Put n = 2

a₂ = - 4(2)² + 15 = -1

Put n = 3

a₃ = - 4(3)² + 15 = -21

Common difference, d₁= a₂ – a₁ = -1 – 11 = -12

Common difference, d₂= a₃ – a₂ = - 21 + 1 = 20

Since, d₁ ≠ d₂

Therefore, it’s not an A.P

S_n = 3n² + 5n

Taking n = 1, we get

S₁ = 3(1)² + 5(1)

⇒ S₁ = 3 + 5

⇒ S₁ = 8

⇒ a₁ = 8

Taking n = 2, we get

S₂ = 3(2)² + 5(2)

⇒ S₂ = 12 + 10

⇒ S₂ = 22

∴ a₂ = S₂ – S₁ = 22 – 8 = 14

Taking n = 3, we get

S₃ = 3(3)² + 5(3)

⇒ S₃ = 27 + 15

⇒ S₃ = 42

∴ a₃ = S₃ – S₂ = 42 – 22 = 20

So, a = 8,

d = a₂ – a₁ = 14 – 8 = 6

Now, we have to find the 15^th term

a_n = a + (n – 1)d

a₁₆ = 8 + (16 – 1)6

a₁₆ = 8 + 15 × 6

a₁₆ = 8 + 90

a₁₆ = 98

Hence, the 16^th term is 98.

Correct option is (D) 1 + 2 + 3 + …………

(A) Given sequence is 1, 8, 27, .....

\(\because\) \(a_2-a_1\) = 8 - 1 = 7,

\(a_3-a_2\) = 27- 8 = 19

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

\(\therefore\) Given sequence is not an arithmetic progression.

(B) Given sequence is 1, 4, 9, .......

\(\therefore\) \(a_2-a_1\) = 4 - 1 = 3

& \(a_3-a_2\) = 9 - 4 = 5

\(\because\) \(3\neq5\)

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

\(\therefore\) Given sequence is not an arithmetic progression.

(C) Given sequence is 1, -2, 3, -4, .......

\(\therefore\) \(a_2-a_1\) = -2 - 1 = -3,

\(a_3-a_2\) = 3 - (-2)

= 3+2 = 5

\(\because\) \(-3\neq5\)

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

\(\therefore\) Given sequence is not an arithmetic progression.

(D) Given sequence is 1, 2, 3, ......

\(\therefore\) \(a_2-a_1\) = 2 - 1 = 1,

\(a_3-a_2\) = 3 - 2 = 1

\(\because\) \(a_2-a_1\) \(=a_3-a_2\)

\(\therefore\) Given sequence is an arithmetic progression (A.P.).

Correct option is D) 1 + 2 + 3 + …………

Given:

The sequence a_n = 3n² - 5.

To prove:

the sequence defined by a_n = 3n² - 5 is not A.P.

Proof:

Consider the sequence a_n = 3n² - 5,

Put n = 1

a₁= 3(1)² – 5 = 3 – 5 = -2

Put n = 2

a₂= 3(2)² – 5 = 12- 5 = 7

Put n = 3

a₃= 3(3)² – 5 = 27 – 5 = 22

In an A.P the difference of consecutive terms should be same.

So,

Common difference, d₁= a₂ – a₁ =7 – (-2) = 9

Common difference, d₂ = a₃ – a₂ = 22 – 9 = 13

Since, d₁ ≠ d₂

Therefore, it’s not an A.P.

Correct option is B) 5/9

We know that nth term of an A.P is given by,

a_n = a + (n – 1) d

Now equating it with the expression given we get,

2 n + 1 = a + (n – 1) d

2 n + 1 = a + nd – d

2 n + 1 = nd + (a – d)

Equating both sides we get,

d = 2 and a – d = 1

So we get,

a = 3 and d = 2.

So the first term of this sequence is 3, and the common difference is 2.

Correct option is (C) 5

Given sequence is \(\sqrt3 ,\sqrt{12} ,\sqrt{27},........\)

\(\Rightarrow\sqrt{3},\sqrt{4\times3},\sqrt{9\times3},.....\) is the given sequence

\(\Rightarrow\) \(\sqrt3,2\sqrt{3},3\sqrt{3},......\) is the given sequence

\(\because\) \(1^{st}\) term of sequence \(=\sqrt3\)

\(2^{nd}\) term of sequence \(=2\sqrt3\)

\(3^{rd}\) term of sequence \(=3\sqrt3\)

 \(\vdots\)

\(10^{th}\) term of sequence \(=10\sqrt3=\sqrt{3\times10^2}\)

\(\therefore n=10\)    (By comparing \(\sqrt{3n^2}\) with \(\sqrt{3\times10^2})\)

\(\Rightarrow\) \(\frac{n}{2}=\frac{10}{2}=5\)

Correct option is C) 5

Correct option is (C) 2

\(n^{th}\) term of sequence \(=\frac{3n-4}{7}\)

i.e., \(a_n=\frac{3n-4}{7}\)

Now, \(6^{th}\) term of the sequence is

\(a_6=\frac{3\times6-4}7=\frac{18-4}7\)

\(=\frac{14}7=2\)

Correct option is C) 2

First n odd numbers are 1, 3, 5, 7, 9........., (2n–1) which forms an A.P a= 1 

l= 2n – 1 where l be the last term 

S_n = \(\frac{n}{2}\)(2a + (n–1) d) or \(\frac{n}{2}\)(a + l) 

= \(\frac{n}{2}\)(1 + 2n –1) 

= \(\frac{n}{2}\)(2n) 

=n (n) 

=n²

First n even Natural numbers are 2, 4, 6,……, 2n 

This forms an A.P where 

a= 2

l= 2n 

where l is last term of the A.P 

S_n = \(\frac{n}{2}\)(2a + (n–1) d) or \(\frac{n}{2}\)(a + l)

= \(\frac{n}{2}\)(2 + 2n)

= \(\frac{n}{2}\)[2(1 + n)]

= n (n + 1)

Given,

First term, a = -12

Common difference, d = a₂ – a₁ = – 9 – (- 12)

d = – 9 + 12 = 3

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

⟹ 21 = -12 + (n – 1)3

⟹ 21 = -12 + 3n – 3

⟹ 21 = 3n – 15

⟹ 36 = 3n

⟹  n = 12

Thus, the number of terms is 12.

Now, if 1 is added to each of the 12 terms, the sum will increase by 12.

Hence, the sum of all the terms of the A.P. so obtained is

⟹ S₁₂ + 12  = 12/2[a + l] + 12

= 6[-12 + 21] + 12

= 6 × 9 + 12

= 66

Therefore, the sum after adding 1 to each of the terms in the A.P is 66.

Given an A.P in which a₄ + a₈ = 24 

⇒ a + 3d + a + 7d = 24 

⇒ 2a + 10d = 24 

⇒ a + 5d = 12 ……. (1) 

and a₆ + a₁₀ = 44 

⇒ a + 5d + a + 9d = 44 

⇒ 2a + 14d = 44 

⇒ a + 7d = 22 ……. (2) 

Also a + 5d = 12 

⇒ a + 5(5) = 12 

⇒ a + 25 = 12 

⇒ a = 12 – 25 = -13 

∴ The A.P is a, a + d, a + 2d, …… 

i.e., – 13, (- 13 + 5), (-13 + 2 × 5)… 

⇒ -13, -8, -3, …….

Given: An A.P: 3, 8, 13, …… , 253 

Here a = a₁ = 3 

d = a₂ – a₁ = 8 – 3 = 5 

a_n = 253, where 253 is the last term 

a_n = a + (n – 1)d 

∴ 253 = 3 + (n – 1)5 

⇒ 253 = 3 + 5n – 5 

⇒ 5n = 253 + 2 

⇒ n = 255/5 = 51 

∴ The 20^th term from the other end would be 

1 + (51 – 20) = 31 + 1 = 32 

∴ a₃₂ = 3 + (32 – 1) × 5 

= 3 + 31 × 5 

= 3 + 155 = 158

4n+3 because n^th term of an AP can only be a linear relation in n as a_n=a+(n-1)d

a_n is the n^th term of an A.P and a_n–1 is the (n–1)^th term of an A.P, 

d = common difference, 

S_n = sum of n terms of an A.P 

d= a_n – a_n–1 

But a_n= S_n – S_n–1 

And a_n–1= S_n–1 – S_n–2 

So d= S_n – S_n–1 – (S_n–1 – S_n–2) 

d= S_{n – 2} S_n–1 + S_n–2

Correct option is (A) \(a_n = a+(n-1)d\)

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

Correct option is A) a_n = a + (n – 1)d 

Correct option is (D) a – 3d

Let \(a_1=a+3d,\)

\(a_2=a+d\;\&\;a_3=a-d\)

\(\therefore d=a_2-a_1\)

\(=(a+d)-(a+3d)=-2d\)

\(a_4=a_3+d\)

\(=(a-d)+(-2d)\)

\(=a-3d\)

Correct option is D) a – 3d

Correct option is (A) 24

Given that 16, x, 36 are in G.P.

\(\therefore b^2=ac\)

\(\Rightarrow x^2=16\times36\)

\(=4^2\times6^2\)

\(=(4\times6)^2\)

\(=24^2\)

\(\Rightarrow x=24\)

Correct option is A) 24

Correct option is (C) t₁₁

Given A.P. is -18, -16, -14, ….....

\(\because a_1=-18,a_2=-16\)

\(d=a_2-a_1\)

= -16 - (-18)

= -16+18 = 2

\(\because t_n=a+(n-1)d\)           \((\because a_1=a)\)

First positive term in given A.P. is 2.

\(\therefore2=-18+(n-1)(2)\)

\(\Rightarrow2+18=2(n-1)\)

\(\Rightarrow n-1=\frac{20}2=10\)

\(\Rightarrow n=10+1=11\)

Hence, \(11^{th}\) term of A.P. is first positive term.

Correct option is C) t₁₁

Solution:

We have First odd number is 1
Last odd number is 49
a = 1 an or I = 49 ,common difference d= 2.
a_n = a+(n-1)d
=1+(n-1)2 = 49
=1+2n-2 =49
=2n-1 =49
=2n=49+1
=2n=50
n=25
s_n=n/2(1+49)
=25/2(1+49)
=25/2* 50
25 *25 = 625
therefore sum of all odd numbers between 0 to 50 is 625.

Correct option is (B) \(\sqrt{ac} \)

\(\because\) a, b, c are in G.P.

\(\therefore b^2=ac\)

\(\Rightarrow\) \(b=\sqrt{ac} \)

Correct option is B) \(\sqrt{ac}\)

Correct option is (D) 5050

First natural numbers is 1.

\(\therefore a=1\)

\(100^{th}\) natural numbers is \(100=l\)   (Let)

\(\therefore S_n=\frac n2(a+l)\)

\(=\frac{100}2(1+100)\)     \((\because a=1,n=100\;\&\;l=100)\)

\(=50(101)\) = 5050

Correct option is D) 5050

Here 1^st term x = 72 and common difference d = 70 - 72 = - 2

So,  For finding the value of n

a_n = a + (n - 1)d

=> 40 = 72 + (n - 1) (-2) 

=>40 - 72 = - 2n + 2

=> -32 = - 2n + 2 

=> -34 = - 2n

=> n = 17 

So, 17^th term is 40.

Given : a₃ = a + (3 - 1) d = a + 2d = 16 .....(i)

a₇ - a₅ = 12 ....(ii)

(a + 6d) - (a + 4d) = 12

a + 6d - a - 4d = 12     2d = 12

d = 6 

Put d = 6 in equation (i)

a = 16 – 12 

a = 4 

So, A.P. is 4, 10, 16, 22, 28, ......

Those sequence whose terms follow certain patterns are called progression.

Generally there are three types of progression.

(i) Arithmetic Progression (A.P.) (ii) Geometric Progression (G.P.) (iii) Harmonic Progression (H.P.)

Given A.P : 0.6, 1.7, 2.8,…. S₁₀₀

a = 0.6; d = a₂ – a₁ = 1.7 – 0.6 = 1.1; n = 100 

S_n = \(\frac{n}{2}\) [2a + (n – 1)d] 

∴ S₁₀₀ = \(\frac{100 }{2}\) [2 × 0.6 + (100 – 1)1.1] 

= 50 [1.2 + 99 × 1.1] 

= 50 [1.2 + 108.9] 

= 50 × 110.1 

= 5505

Correct option is (B) 3, 5, 7

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

Given that their sum is 15.

\(\therefore\) (a - d) + a + (a+d) = 15

\(\Rightarrow3a=15\)

\(\Rightarrow a=\frac{15}3=5\)

Also given that the sum of the squares of the extremes is 58.

\(\therefore(a-d)^2+(a+d)^2=58\)

\(\Rightarrow(5-d)^2+(5+d)^2=58\)    \((\because a=5)\)

\(\Rightarrow 2(5^2+d^2)=58\)            \((\because(a-b)^2+(a+b)^2=2(a^2+b^2))\)

\(\Rightarrow5^2+d^2=\frac{58}2=29\)

\(\Rightarrow d^2=29-5^2=29-25\)

\(=4=2^2\)

\(\Rightarrow d=2\)

\(\therefore a-d=5-2=3\)

& \(a+d=5+2=7\)

Hence, the required numbers are 3, 5 and 7.

Correct option is B) 3, 5, 7

Given, 

First term, a = 2 

Last term, a_n = l = 29 

And, Sum of terms, S_n = \(\frac{n}{2}\)[a + l]

155 = \(\frac{n}{2}\)[ 2 + 29]

155 x 2 = n x 31

n = \(\frac{155 \times 2}{31}\) = \(\frac{310}{31} = 10\)

We know, a_n = a + (n - 1)d

 29 = 2 = (10 - 1)d

29 - 2 = 9d

27 = 9d

d = 3

Hence, common difference, d = 3

If 45, a and 2 are three consecutive terms of an AP, then we have:

a -45 = 2-a

2a=2+45

2a=47

a = 24

First n natural numbers: 1, 2, 3, 4, 5, …, n

Here, a = 1, d = 1

Sum = S_n = n/2 [2a + (n-1)d]

= n/2 [2(1) + (n-1)(1)]

= n(n+1)/2

First n even natural numbers: 2, 4, 6, 8, 10,….,n

Here, a = 2, d = 4 – 2 = 2

Sum = S_n = n/2 [2a + (n-1)d]

= n/2 [2(2) + (n-1)(2)]

= n(n+1)

Let (2p-1), 7 and 3p be three consecutive terms of an AP.

Then 7-(2p-1) = 3p -7

5p = 15

p = 3

When p = 3, (2p-1), 7 and 3p form three consecutive terms of an AP.

Given:

First term =a = p and

Common difference = d = q

Now,

a₁₀ = a + (n – 1) d

= p + (10 – 1)q

= (p + 9q)

Given:

a₅ = 22

a + 4d = 22 (i)

Acc. to question,

a₉ = 6a₂ (ii)

a₉ = a +8d

a₂ = a+d

Putting the value of a₉ and a₂ in (ii)

a +8d = 6(a +d)

2d = 5a

2d = 5(22 - 4d)

22d = 110

d = 5

Now, putting the value of d in (i)

a = 22 - 4(5) = 2

a₁ = a = 2

a₂ = a +d = 2 +5 = 7

a₃ = a +2d = 2 + 2(5) = 12

Hence, the A.P. is 2, 7, 12, 17,..........

Given, (2p – 1), 7, 3p are in AP

Then,

7 – (2p – 1) = 3p – 7

7 – 2p + 1 = 3p – 7

5p = 15

p = 3

The value of p is 3.

AP terms: 4/5, a, 2 (given)

Then,

a – 4/5 = 2 – a

a = 7/5

a₆ = a + 5d

12 = a + 5d …(i)

a₈ = a + 7d

22 = a + 7d …(ii)

Subtracting (i) from (ii), we get,

10 = 2d

d = 5

from (i)

12 = a + 5(5)

12 = a + 25

a = -13

a₂ = a +d

= -13 +5

= - 8

a_n = a + (n -1)d

= - 13 + (n - 1)5

= -18 + 5n

Let (2p - 1), 7 and p be three consecutive terms of an AP.

Then 7-(2p - 1) = 3p - 7

\(\Rightarrow\) 5p = 15

\(\Rightarrow\) p = 3

∴ When p = 3,(2p - 1), 7 and 3p form three consecutive terms of an AP.

Let (2p + 1), 13, (5p - 3) be three consecutive terms of an AP.

Then 13 - (2p + 1) = (5p - 3) - 13

\(\Rightarrow\) 7p = 28

\(\Rightarrow\) p = 4

∴ When p = 4,(2p + 1),13 and (5p - 3) from three consecutive terms of an AP.

Given: The first term of an A.P. is p and its common difference is q. 

To find: its 10^th term. 

Solution: The first term of A.P is p. Common difference = q 

We know an = a + (n–1)d Where a is first term and d is common difference. 

So,a₁₀ = p + (10-1)q 

a₁₀ = p + 9q

If \(\frac{4}{5}\),a, 2 are three terms in an A.P. so there common difference will be same 

a - 4/5 = 2 - a

2a = \(\frac{14}{5}\)

a = \(\frac{7}{5}\)

If a,b,c are three consecutive terms of an AP

Then,

 \(b = {c + a \over 2}\)

Thus  \(a= {4/5 +2 \over 2}\)

Ans: a=1.4 or 7/5

Given, 2p + 1, 13, 5p – 3 are in AP

Then,

13 – (2p + 1) = (5p – 3) – 13

13 – 2p – 1 = 5p – 3 – 13

12 – 2p = 5p – 16

p = 4

The value of p is 4.

Given: 2p + 1, 13, 5p – 3 are three consecutive terms of an A.P. 

To find: The value of p 

Solution: Consider a₁ = 2p + 1 , a₂ = 13 and a₃ = 5p – 3

The A.P will be of the form 2p + 1, 13, 5p – 3,.........

Since the terms are in A.P so the common differences in them is same.

⇒ a₂ - a₁ = a₃ - a₂ 

⇒ 13– (2p + 1) = 5p –3 –13

⇒ 13– 2p - 1 = 5p –3 –13 

⇒12– 2p = 5p –16 

⇒12+16 = 5p +2p 

⇒ 28 = 7p

⇒ p= 4

Let (2p+1), 13, (5p-3) be three consecutive terms of an AP.

Then 13-(2p+1) = (5p-3) -13

7p = 28

p = 4

When p = 4, (2p+1), 13 and (5p-3) from three consecutive terms of an AP.

Correct option is (B) 2

\(S_n=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.......+\) upto n terms

\(=\cfrac{1(1-(\frac{1}{2})^n)}{1-\frac{1}{2}}\)      \((\because S_n=\frac{a(1-r^n)}{r-1},r<1\) in G.P.)

\(=\cfrac{1-\frac{1}{2^n}}{\frac{1}{2}}\)

\(=2(1-\frac1{2^n})\)

Sum of infinite terms \(=S_\infty=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......(\infty-terms)\)

\(=\underset{n\rightarrow\infty}{lim}\,S_n\)

\(=\underset{n\rightarrow\infty}{lim}\,2(1-\frac1{2^n})\)

\(=2(1-\frac1\infty)\)

\(=2(1-0)=2\)

Correct option is B) 2