JEE Advanced (Single Correct MCQs): Sequences and Series - JEE MCQ

∵ x, y, z are the p^th, q^th and r^th terms of an  AP..
∴ x = A + (p – 1)D; y = A + (q – 1) D; z  = A + (r – 1) D
⇒ x – y = (p – q)D;  y – z = (q – r) D z – x = (r – p) D ....(1)
where A is the first term and D is the common difference.
Also x, y, z are the p^th, q^th, and r^th terms of a GP.
∴ x = AR^{p – 1}, y = AR^{q – 1}, z = AR^{r – 1}
∴ x^{y – z} y^{z – x} z^{x –y} = (AR^{p –1})^y–z (AR^q–1)^z–x (AR^r–1)^x–y
= A^{y–z+z–x+x–y}R^{(p–1)(y–z)+(q–1)(z–x)+(r–1)(x–y)}
= A⁰r^{(p–1) (q–r) D+(q–1)(r–p)D+(r–1)(p–q)D} = A⁰R⁰ =1

ar² = 4 ....(1)

a. ar. ar² .ar³ .ar⁴ = a⁵ r¹⁰ = (ar²)⁵ = 45.

= 2 + .357 + 0.000357 + ...∞

a, b, c are in G.P. b² = ac ....(1)

ax²  + 2bx + c = 0 and dx²  + 2ex + f = 0 have a common root

Let it be α, then aα² + 2bα + c = 0 dα² + 2eα + f = 0

⇒

⇒

Substituting the value of α , we get

⇒ (cd – af) ² = 4 (ae – bd) (bf – ce)

Dividing both sides by a²c² we get

[Using eq. (1)]

⇒

⇒

[Using eq.(1)]

⇒

⇒  are in A.P..

Let S n terms

..n terms

= (1 + 1 + 1 + ....n  terms) 

=    = n - 1+2^-n

We know that log₂ 4 = 2 and log₂ 8 = 3
∴ log₂ 7 lies between 2 and 3
∴ log₂ 7 is either rational or irrational but not integer or prime number.
If possible let log₂   (a rational number)

⇒ 2^p/q = 7
⇒ 2^p = 7^q
⇒ even number = odd number
∴ We get a contradiction, so assumption is wrong.
Hence log₂ 7 must be an irrational number.

In (a + c), In (a – c), In (a – 2b + c) are in A.P.
⇒ a + c, a – c, a – 2b + c are in G.P.
⇒ (c – a)² = (a + c) (a –2b + c)
⇒ (c – a)² = (a + c)² – 2b (a + c)
⇒ 2b (a + c) = (a + c)2 – (c – a)²
⇒ 2b (a + c) = 4ac ⇒ 

⇒ a, b, c  are in H.P

a₁= h₁= 2, a₁₀ = h₁₀ = 3
3 = a₁₀ = 2 + 9d ⇒ d = 1/9
∴ a₄ = 2 + 3d = 7/3

∴ H = 4.

Sum = 4 and second term = 3/4, it is given that first term is a and common ratio r

⇒ = 4 and ar = 3 /4 ⇒ 

Therefore, 

or a² – 4a + 3 = 0 ⇒ (a – 1)  (a – 3) = 0
⇒ a = 1  or  3
When a = 1, r = 3/4 and when a = 3, r = 1/4

α, β are the roots of x² – x + p=0
∴ α + β = 1 ....(1)
αβ = p ...(2)
g,δ are the roots of x² – 4x  + q = 0

∴ γ + δ = 4....(3)
γδ = q ....(4)
α,β, γ,δ are in G.P..
∴   Let α = a;  β = ar, γ = ar ² , δ = ar³ .
Substituting these values in equations (1), (2), (3) and (4), we get

a + ar = 1 ....(5)
a²r = p ....(6)
ar² + ar³ = 4 ....(7)
a²r⁵ = q ....(8)
Dividing (7)  by  (5)  we get

⇒ 

(5) ⇒  or 

As p is an integer (given), r is also an integer (2 or – 2).

∴ (6) ⇒ Hence a = -1 and r = -2.

∴ p = (– 1)² × (– 2) = – 2

q = (– 1)² × (– 2)⁵ = – 32

a, b, c, d are in A.P.

∴ d, c, b, a are also in A.P.

⇒ are also in A.P..

⇒ are in A.P..

⇒ abc, abd, acd, bcd are in H.P.

ATQ  2 + 5 + 8 + .... 2n terms = 57 + 59 + 61 + .... n terms

⇒

⇒ 6n + 1= n + 56 ⇒ 5n = 55 ⇒ n = 11

Given that a, b, c are in A.P.
⇒ 2b = a + c
⇒ but given a + b + c = 3/2  
⇒  3b = 3/2
⇒ b = 1/2 and then a + c = 1
Again a², b², c², are in G.P. ⇒ b⁴ = a² c²

⇒ 

and a + c = 1 ....(1) Considering a + c = 1 and ac = 1/4
⇒ (a – c)² = 1 –  1 = 0   ⇒ a = c  
but a ≠ c as given that a < b < c
∴ We consider a + c = 1 and ac = – 1/4
⇒ (a – c)² = 1 + 1= 2 ⇒ 

but a < c ⇒                 ....(2)

Solving (1) and (2) we get 

Since G.P. contains infinite terms

∴ – 1 < r < 1

⇒ 0 < x< 10.

In the quadratic equation ax² + bx + c = 0

Δ = b² - 4ac and 

and 

Given are in G.P.

⇒ are in G.P.

⇒ 

⇒ b⁴ + 4a²c² – 4ab²c = b⁴ – 3ab²c

⇒ 4a²c² – ab²c = 0 ⇒ ac Δ = 0

Given that for an A.P.S_n = cn²

Then T_n = S_n-S_n-1 = cn² - c(n- 1)² = (2n- 1)c

∴ Sum of squares of n terms of this A.P

∵ a₁, a₂, a₃, ....... are in H.P..

.. are in A.P..

Now  

Clearly  a_n < 0  if 

⇒ – 4n < – 99   or  

Hence least value of n is 25.

log_c b₁, log_c b₂,----, log_cb₁₀₁ are in A.P.
⇒ b₁, b₂,---- b₁₀₁ are in G.P.
Also a₁, a₂,---- a₁₀₁ are in A.P. where a₁ = b₁ are a₅₁ = b₅₁.
∴ b₂, b₃,---,b₅₀ an d GM’s and a₂, a₃,----, a₅₀ are AM’s between b₁ and b₅₁.
∵ GM < AM ⇒ b₂ < a₂, b₃ < a₃,----, b₅₀ < a₅₀
∵ b₁ + b₂ +----+ b₅₁ <  a₁ + a₂ +----+ a₅₁
⇒ t < s Also a₁, a₅₁, a₁₀₁ are in AP b₁, b₅₁, b₁₀₁ are in GP
∵ a₁ = b₁ and a₅₁ = b₅₁
∴ b₁₀₁ > a₁₀₁