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Test: Single Correct MCQs: Sequences and Series | JEE Advanced

19 Questions MCQ Test Mathematics For JEE | Test: Single Correct MCQs: Sequences and Series | JEE Advanced

Description

QUESTION: 1

If x, y and z are pth, qth and rth terms respectively of an A.P. and also of a G.P., then x^y – ^z y^z – ^x z^{x – y} is equal to :

A.
xyz
B.
0
C.
1
D.
None of these

Solution:

∵ 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

QUESTION: 2

The third term of a geometric progression is 4. The product of the first five terms is (1982 - 2 Marks)

A.
4³
B.
4⁵
C.
4⁴
D.
none of these

Solution:

ar² = 4 ....(1)

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

QUESTION: 3

The rational number, which equals the number with recurring decimal is (1983 - 1 Mark)

A.
B.
C.
D.
none of these

Solution:

= 2 + .357 + 0.000357 + ...∞

QUESTION: 4

If a, b, c are in G.P., then the equations ax²+ 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if are in –– (1985 - 2 Marks)

A.
A.P.
B.
G.P.
C.
H.P.
D.
none of these

Solution:

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..

QUESTION: 5

Sum of the first n terms of the series....... is equal to (1988 - 2 Marks)

A.
2ⁿ - n-1
B.
1 -2^{- n}
C.
n + 2^{- n} – 1
D.
2ⁿ + 1.

Solution:

Let S n terms

..n terms

= (1 + 1 + 1 + ....n terms)

= = n - 1+2^-n

QUESTION: 6

The number log₂ 7 is (1990 - 2 Marks)

A.
an integer
B.
a rational number
C.
an irrational number
D.
a prime number

Solution:

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.

QUESTION: 7

If ln(a + c), ln (a – c), ln (a – 2b + c) are in A.P., then (1994)

A.
a, b, c are in A.P.
B.
a², b², c² are in A.P.
C.
a, b, c are in G.P.
D.
a, b, c are in H.P.

Solution:

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

QUESTION: 8

Let a₁, a₂, ..... a₁₀be in A, P, and h₁, h₂,....h₁₀ be in H.P. If a₁ = h₁ = 2 and a₁₀ = h₁₀= 3, then a₄h₇ is (1999 - 2 Marks)

A.
2
B.
3
C.
5
D.
6

Solution:

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

QUESTION: 9

The harmonic mean of the roots of the equation is (1999 - 2 Marks)

A.
2
B.
4
C.
6
D.
8

Solution:

∴ H = 4.

QUESTION: 10

Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 3/4, then (2000S)

Solution:

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

QUESTION: 11

Let α, β be the roots of x² - x + p = 0 and γ, δ be the roots of x²- 4x + q = 0. If α, β, γ, δ are in G.P., then the integral values of p and q respectively, are (2001S)

A.
–2, –32
B.
–2, 3
C.
–6, 3
D.
–6, –32

Solution:

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 = ar² , 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

QUESTION: 12

Let the positive numbers a, b, c, d be in A.P. Then abc, abd, acd, bcd are (2001S)

A.
NOT in A.P./G.P./H.P.
B.
in A.P.
C.
in G.P.
D.
in H.P.

Solution:

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.

QUESTION: 13

If the sum of the first 2n terms of the A.P. 2, 5, 8, …, is equal to the sum of the first n terms of the A.P. 57, 59, 61, …, then n equals(2001S)

A.
10
B.
12
C.
11
D.
13

Solution:

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

⇒

⇒ 6n + 1= n + 56 ⇒ 5n = 55 ⇒ n = 11

QUESTION: 14

Suppose a, b, c are in A.P. and a², b², c²are in G.P. if a < b < c and a + b + c =, then the value of a is (2002S)

Solution:

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

QUESTION: 15

An infinite G.P. has first term ‘x’ and sum ‘5’, then x belongs to (2004S)

A.
x < – 10
B.
– 10 < x < 0
C.
0 < x < 10
D.
x > 10

Solution:

Since G.P. contains infinite terms

∴ – 1 < r < 1

⇒ 0 < x< 10.

QUESTION: 16

In the quadratic equation ax² + bx + c = 0, Δ = b² – 4ac and α + β, α² + β², α³+ β³, are in G.P. where a, b are the root of ax²+ bx + c = 0, then (2005S)

A.
Δ ≠ 0
B.
bΔ = 0
C.
cΔ = 0
D.
Δ = 0

Solution:

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

QUESTION: 17

In the sum of first n terms of an A.P. is cn², then the sum of squares of these n terms is (2009)

Solution:

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

QUESTION: 18

Let a₁, a₂, a₃, ..... be in harmonic progression with a₁ = 5 and a₂₀ = 25. The least positive integer n for which an < 0 is (2012 )

A.
22
B.
23
C.
24
D.
25

Solution:

∵ 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.

QUESTION: 19

Let b_i > 1 for i = 1, 2, ..., 101. Suppose loge b₁, loge b₂, ...., log_e b₁₀₁ are in Arithmetic Progression (A.P.) with the common difference log_e2. Suppose a₁, a₂, ...., a₁₀₁ are in A.P. such that a₁= b₁ and a₅₁= b₅₁. If t= b₁+b₂ + .... + b₅₁ and s= a₁+a₂+ .... + a₅₃, then (JEE Adv. 2016)

A.
s > t and a₁₀₁> b₁₀₁
B.
s > t and a₁₀₁ < b₁₀₁
C.
s < t and a₁₀₁ > b₁₀₁
D.
s < t and a₁₀₁< b₁₀₁

Solution:

log_c b₁, log_cb₂,----, 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 = b₁and a₅₁ = b₅₁
∴ b₁₀₁ > a₁₀₁