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Correct option is (C) -57

First term of A.P is a = -2,

Common difference of A.P. is d = -5

\(\therefore a_{12}=a+11d\)

\(=-2+11\times-5\)

\(=-2-55=-57\)

\(\therefore\) \(12^{th}\) term is -57.

Correct option is C) -57

Correct option is (B) 5/7

\(n^{th}\) term of A.P. is \(a_n=\frac{5n-4}{7}\)

\(\therefore\) Common difference is \(d=a_n-a_{n-1}\)

\(=\frac{5n-4}{7}-\frac{5(n-1)-4}{7}\)

\(=\frac{5n-4-5n+5+4}{7}\) \(=\frac{5}{7}\)

Correct option is B) 5/7

Correct  answer is A. 5,10,15,20

Let the 4 numbers be a, a + d, a + 2d, a + 3d. 

Sum of 4 numbers in A.P = 50 

a + a + d + a + 2d + a + 3d = 50 

⇒ 4a + 6d = 50 

⇒ 2a + 3d = 25 --------------(1) 

Given the greatest number is 4 times the least. 

4(a) = a + 3d 

4a – a = 3d 

a = d

Putting, a = d in (1), we obtain 

5d = 25 

d = 5 

⇒ a = 5 

∴ First four terms are 5, 10, 15, 20.

Let’s consider the four terms of the A.P. to be (a – 3d), (a – d), (a + d) and (a + 3d).

From the question,

Sum of these terms = 50

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 50

⇒ a – 3d + a – d + a + d + a – 3d= 50

⇒ 4a = 50

⇒ a = 50/4 = 25/2

And, also given that the greatest number = 4 x least number

⇒ a + 3d = 4 (a – 3d)

⇒ a + 3d = 4a – 12d

⇒ 4a – a = 3d + 12d

⇒3a = 15d

⇒a = 5d

Using the value of a in the above equation, we have

⇒25/2 = 5d

⇒ d = 5/2

So, the terms will be:

(a – 3d) = (25/2 – 3(5/2)), (a – d) = (25/2 – 5/2), (25/2 + 5/2) and (25/2 + 3(5/2)).

⇒ 5, 10, 15, 20

Option : (A)

Let 4 numbers of A.P. be the a, a+d, a+2d, a+3d. 

Here, 

Given that their sum is 50. 

i.e., 

a+a+d+a+2d+a+3d = 50 

∴ 4a + 6d = 50 

∴ 2a + 3d = 25 ....(1) 

Also, 

Given that,

The greatest number is 4 times the least. 

i.e.,

4a = a + 3d 

∴ 3a = 3d 

∴ a = d 

Substituting a=d in (1) we get, 

5a = 25 

∴ a = 5 

And, 

d = a = 5 

So, numbers are a = 5; 

a+d = 5+5; 

a+2d = 5+2(5); 

a+3d = 5+3(5) 

i.e. 5,10,15,20

a = 4, d = -3

Given, first term (a) = 4

Common difference (d) = -3

Then arithmetic progression is: a, a + d, a + 2d, a + 3d, ……

⇒ 4, 4 – 3, a + 2(-3), 4 + 3(-3), ……

⇒ 4, 1, – 2, – 5, – 8 ……..

Given First term (a) = –1.5

Common difference (d) = – 0 5

Then arithmetic progression is; a, a + d, a + 2d, a + 3d, ……

⇒ -1.5, -1.5, -0.5, –1.5 + 2(– 0.5), –1.5 + 3(– 0.5)

⇒ – 1.5, – 2, – 2.5, – 3, …….

Given, first term (a) = -1

Common difference (d) = \(\frac{1}{2}\)

Then arithmetic progression is: a, a + d, a + 2d, a + 3d,

⇒ -1, -1 + \(\frac{1}{2}\), -1, 2\(\frac{1}{2}\), -1 + 3\(\frac{1}{2}\), …

⇒ -1, -\(\frac{1}{2}\), 0, \(\frac{1}{2}\)

Given: AP is 41, 38, 35,…, 8

Here, first term = a = 41

Last term = 8

Common difference = d = 38 – 41 = -3

To find: the number of terms (n)

Last term = a + (n – 1)d

8 = 41 + (n – 1)(-3)

8 – 41 = (n – 1)(-3)

n – 1 = 11

n = 11 + 1 = 12

There are 12 terms.

Here, a = 41 , d = 38 – 41 = –3 and l = 8

where l = a + (n – 1)d

⇒ 8 = 41 + (n – 1)(–3)

⇒ 8 = 41 –3n + 3

⇒ 8 = 44 – 3n

⇒ 8 – 44 = –3n

⇒ –36 = –3n

⇒ n = -36/-3 = 12

Hence, the number of terms in the given AP is 12

AP = 5, 8, 11, 14, …

Here, a = 5 and d = 8 – 5 = 3

Let 51 be a term, say, nth term of this AP.

We know that

a_n = a + (n – 1)d

So, 51 = 5 + (n – 1)(3)

⇒ 51 = 5 + 3n – 3

⇒ 51 = 2 + 3n

⇒ 51 – 2 = 3n

⇒ 49 = 3n

⇒ n = 49/3

But n should be a positive integer because n is the number of terms. So, 51 is not a term of this given AP.

AP = 999, 995, 991, 987,…

Here, a = 999, d = 995 – 999 = –4

a_n < 0

⇒ a + (n – 1)d < 0

⇒ 999 + (n – 1)(–4) < 0

⇒ 999 – 4n + 4 < 0

⇒ 1003 – 4n < 0

⇒ 1003 < 4n

⇒  1003/4 < n

⇒ n > 250.75

Nearest term greater than 250.75 is 251

So, 251^st term is the first negative term

Now, we will find the 251^st term

a_n = a +(n – 1)d

⇒ a₂₅₁ = 999 + (251 – 1)(–4)

⇒ a₂₅₁ = 999 + 250 × –4

⇒ a₂₅₁ = 999 – 1000

⇒ a₂₅₁ = – 1

∴, –1 is the first negative term of the given AP.

Given : 

n^th term of the A.P. 9, 7, 5,… is same as the nth term of the A.P. 15, 12, 9 … 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a1 and d is common difference and n is any natural number 

Let A.P. 9, 7, 5,… has first term a₁ and common difference d₁ 

⇒ a₁ = 9 and a₂ = 7 

Common difference, 

d₁ = a₂ – a₁ 

= 7 – 9 

= -2 

Now, 

a_n = a₁ + (n – 1)d₁ 

⇒ a_n = 9 + (n – 1)(-2) 

⇒ a_n = 9 – 2n + 2 

⇒ a_n = 11 – 2n 

Let A.P. 15, 12, 9 … has first term a₁ and common difference d₁ 

⇒ b₁ = 15 and b₂ = 12 

Common difference, 

d₂ = b₂ – b₁ 

= 12 – 15 

= -3 

Now, 

b_n = b₁ + (n – 1)d₂ 

⇒ b_n = 15 + (n – 1)(-3) 

⇒ b_n = 15 – 3n + 3 

⇒ b_n = 12 – 3n 

According to question : 

a_n = b_n 

⇒ 11 – 2n = 12 – 3n 

⇒ 3n – 2n = 12 – 11 

⇒ n = 1 

Hence,

The value of n is 1.

Given : 

(m + 1)^th term of an A.P. is twice the (n + 1)^th term 

⇒ a_m+1 = 2a_n+1 

To prove : 

a_3m+1 = 2a_m+n+1 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a₁ and d is common difference and n is any natural number 

When n = m + 1 : 

∴ a_m+1 = a + (m + 1 – 1)d 

⇒ a_m+1 = a + md 

When n = n + 1 : 

∴ a_n+1 = a + (n + 1 – 1)d 

⇒ a_n+1 = a + nd 

According to question : 

a_m+1 = 2a_n+1 

⇒ a + md = 2(a + nd) 

⇒ a + md = 2a + 2nd 

⇒ a – 2a + md – 2nd = 0 

⇒ -a + (m – 2n)d = 0 

⇒ a = (m – 2n)d………(i) 

an = a + (n – 1)d 

When n = m + n + 1 : 

∴ a_m+n+1 = a + (m + n + 1 – 1)d 

⇒ a_m+n+1 = a + md + nd 

⇒ a_m+n+1 = (m – 2n)d + md + nd……… (From (i)) 

⇒ a_m+n+1 = md – 2nd + md + nd 

⇒ a_m+n+1 = 2md – nd………(ii) 

When n = 3m + 1 : 

∴ a_3m+1 = a + (3m + 1 – 1)d 

⇒ a_3m+1 = a + 3md 

⇒ a_3m+1 = (m – 2n)d + 3md……… (From (i)) 

⇒ a_3m+1 = md – 2nd + 3md 

⇒ a_3m+1 = 4md – 2nd 

⇒ a_3m+1 = 2(2md – nd) 

⇒ a_3m+1 = 2a_m+n+1…………(From (ii)) 

Hence Proved.

It is given that sum of n terms of AP is nP + 1/2 n(n-1)Q

To find : the common difference 

Substituting n = 1 gives the sum of the first term, that is the first term itself 

a₁ = P 

Substituting n = 2 gives the sum of first two terms 

a₁ + a₂ = 2P + Q 

a₂ = P + Q 

Now common difference d = a₂ - a₁ = Q

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

To Find: Ratio of their 11^th terms

Let us consider 2 AP series AP₁ and AP₂.

Putting n = 1, 2, 3… we get

AP₁ as 8, 15 22… and AP₂ as 31, 35, 39….

So, a₁ = 8, d₁ = 7 and a₂ = 31, d₂ = 4

For AP₁

S₆ = 8 + (11 - 1)7 = 87

For AP₂

S₆ = 31 + (11 - 1)4 = 81

Required ratio = 87/81 = 29/27

It is given that the sum of n terms is 3n² + 2n 

To find : r^th term of an AP, the sum of whose first term is 3n² + 2n 

So, the sum of n - 1 terms is 3(n - 1)² + 2(n - 1) 

Formula, 

T_N = S_N - S_{N - 1} 

So, 

T_r = 3r² + 2r - [3(r - 1)² + 2(r - 1)] 

T_r = 3r² + 2r - [3(r² + 1 - 2r) + 2(r - 1)] 

T_r = - 1 + 6r