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QUESTION: 1

The sum of the series 1^{3} – 2^{3} + 3^{3} – ......+ 9^{3} =

[AIEEE- 2002]

Solution:

(1^{3} + 3^{3} + .....+ 9^{3}) - (2^{3} + 4^{3} + 6^{3}+ 8^{3})

⇒ (1^{3} + 2^{3} + 3^{3} + ... + 9^{3}) - 2.(2^{3} + 4^{3}+6^{3} + 8^{3})

- 2.2^{3} (1 + 2^{3} + 3^{3} +4^{3})

⇒ (45)2 - 16

⇒ (45)^{2} - (40)^{2}

⇒ 85 (5) ⇒ 425

QUESTION: 2

If the sum of an infinite GP is 20 and sum of their square is 100 then common ratio will be =

[AIEEE- 2002]

Solution:

**ANSWER :- c**

**Solution :- **We consider the infinite GP

a,ar,ar^2,..,ar^(n−1),....

We know that, for this GP, the sum of its infinite no. of terms is

S∞ = a/(1−r).

∴ a/(1−r) = 20.........................(1).

The infinite series of which, the terms are the squares of the

terms of the first GP is, a^2+a^2r^2+a^2r^4+... + a^2r^(2n−2)+....

We notice that this is also a Geom. Series, of which the

first term is a^2 and the common ratio r^2.

Hence, the sum of its infinite no. of terms is given by,

S∞ = a^2(1−r^2).

∴ a^2/(1−r^2)= 100........................(2).

(1)÷(2) ⇒(1+r)/a= 1/5......................(3).

Then, (1) × (3) gives ,(1+r)/(1−r) = 4.

⇒ r = 3/5, is the desired common ratio!

QUESTION: 3

If the third term of an A.P. is 7 and its 7th term is 2 more than three times of its 3rd term, then sum of its first 20 terms is-

[AIEEE- 2002]

Solution:

Let AP be a, a + d, a + 2d, .....

∴ T_{3} = a + 2d = 7 ....(i)

& T_{7} = a + 6d = 2 + 3 (7) = 23 .....(ii)

On solving (i) & (ii), we get

d = 4 and a = (–1)

QUESTION: 4

If x_{1}, x_{2}, x_{3} and y_{1}, y_{2}, y_{3} are both in G.P. with the same common ratio, then the points (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})

[AIEEE- 2003]

Solution:

x_{1}, x_{2}, x_{3} and y_{1}, y_{2}, y_{3} are in GP with same common ratio,

∴ (x_{1},y_{1}) ⇒ p(x_{1},y_{1})

(x_{2},y_{2}) ⇒ Q(x_{1}r,y_{1}r)

(x_{3},y_{3}) ⇒ R(x_{1}r^{2},y_{1}r^{2})

∴ P, Q, R are collivear

QUESTION: 5

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation-

[AIEEE- 2004]

Solution:

Let the numbers be a & b,

a + b = 18 & ab = 16

∴ Q.E. with roots a & b is

x^{2}- (a + b)x + ab = 0

⇒ x^{2} - 18x + 16 = 0

QUESTION: 6

Let T_{r} be the r^{th} term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, m ≠ n, T_{m} = 1/n and T_{n} = 1/m, then a - d equals-

[AIEEE- 2004]

Solution:

Given that, T_{m} = 1/n

⇒ a + (m - 1) d = 1/n ......(i)

and T_{n} = 1/m

⇒ a + (n - 1)d = 1/m ......(ii)

On solving Eqs. (i) and (ii), we get

a = d = 1/mn

⇒ a - d = 0

QUESTION: 7

The sum of the first n terms of the series 1^{2} + 2. 2^{2} + 3^{2 }+ 2.4^{2} + 5^{2} + 2.6^{2} +..... is when n is even. When n is odd the sum is-

[AIEEE- 2004]

Solution:

Given that, the sum of n terms of given series is if n is even.

Let n is odd ie, n = 2m + 1

Then, S_{2m+1} = S_{2m} + (2m + 1)th term

QUESTION: 8

where a, b, c are in A.P. and I a I < 1, l b I < 1, I c I < 1 then x, y, z are in -

[AIEEE- 2005]

Solution:

Given that,

** .........**(ii)

Now, a,b,c are in AP.

⇒ -a, -b, -c are in AP.

⇒ 1 - a, 1 - b, 1 - c are also in AP.

are in AP.

⇒ x, y,z are in HP.

QUESTION: 9

If in a ΔABC, the altitudes from the vertices A, B, C on opposite sides are in H.P., then sin A, sin B, sin C are in

[AIEEE- 2005]

Solution:

⇒ a,b,c are in AP

⇒ sin A, sin B, sin C are in AP

QUESTION: 10

Let a_{1}, a_{2}, a_{3}, ..... be terms of an A.P.

[AIEEE- 2006]

Solution:

Where d be a common difference of an AP.

QUESTION: 11

If a_{1}, a_{2}, ..... an are in H.P., then the expression a_{1}a_{2} + a_{2}a_{3} +....+ a_{n –1}an is equal to –

[AIEEE- 2006]

Solution:

Let d be the common difference of AP.

On adding all of these, we get

On putting the value of d in Eq. (i), we get

QUESTION: 12

In a geometric progressionconsisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals-

Solution:

Since, each term is equal to the sum of two preceeding terms.

QUESTION: 13

The sum to infinity of the series

[AIEEE 2009]

Solution:

On subtracting Eq. (ii) from Eq. (i), we get

QUESTION: 14

A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a_{1} = a_{2} = .... = a_{10} = 150 and a_{10}, a_{11} .... are in an AP with common diferencec –2, then the time taken by him to count all notes is -

[AIEEE 2010]

Solution:

Number of notes that the person counts in 10 min

= 10 × 150 = 1500

Since, a_{10}, a_{11}, a_{12}, ..... are in AP with common difference –2.

⇒ Let n be the time taken to count remaining 3000 notes, then

⇒ n^{2} - 149n + 3000 = 0

⇒ (n - 24) (n - 125) = 0

⇒ n = 24, 125

The, the total time taken by the person to count all notes

= 10 + 24 = 34 min.

QUESTION: 15

Statement 1 : The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + ...... ....+(361 + 380 + 400) is 8000.

Statement 2 : for any natural number n.

[AIEEE- 2012]

Solution:

Statement - I : 1+(1+2+4)+(4+ 6+9)+(9+12+16) +..... + (361 + 380 +400)

⇒ 1 + (2^{3}- 1^{3}) + (3^{3} - 2^{3}) + (4^{3}- 3^{3}) + ....+ (20^{3 }- 19^{3})

⇒ (20)^{3} ⇒ 8000

Statement - II : is true and statement - II is correct explanation of Statement - I

QUESTION: 16

If 100 times the 100^{th} term of an AP with non zero common difference equals the 50 times its 50^{ th} term, then the 150^{th} term of this AP is:

[AIEEE 2012]

Solution:

Let the AP be a, a + d, a + 2d, ..... where d ≠ 0

Now, 100 (T_{100}) = 50 (T_{50})

⇒100 (a + 99d) = 50 (a + 49d)

⇒ 50 a = –2450 d ⇒ a = (–149 d)

Now, T_{150} = a + 149 d = 0

QUESTION: 17

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

[JEE 2000,]

Solution:

⇒ 3 = 4r(4-4r)

⇒ 16r^{2} -16r+3 = 0 ⇒ (4r-3)(4r-1) = 0

∴ r = 3/4 & a = 1 or r = 1/4 & a = 3

QUESTION: 18

If a, b, c, d are positive real numbers such that a + b + c + d = 2, then M = (a + b) (c + d) satisifes the relation :

Solution:

a + b + c + d = 2

M is greater than or equal to 0 & less than or equal to 1 or 0 ≤ M ≤ 1

QUESTION: 19

Given that α,γ are roots of the equation, Ax^{2}–4x+1 = 0 and β, δ the roots of the equation, Bx^{2} – 6x + 1 = 0, find values of A and B, such that

[REE 2000, 5]

Solution:

γ satisfy equation (1)

β satisfy equation (2)

QUESTION: 20

Le α, β be the roots of x^{2} – x + p = 0 and γ, δ the roots of x^{2} – 4x + q = 0. If α, β, γ, δ are in G.P., then the integral values of p and q respectively, are

Solution:

α, β, γ, δ in GP.

Let α = a, β = ar, γ = ar^{2}, δ = ar^{3}

a+ar = 1 & ar^{2}+ar^{3 }= 4

a(1+r) = 1 ar^{2} (1 + 1) = 4

r^{2} = 4 = r = ±2

satisfy given equations but p & q not integer for r = 2

⇒ r ≠ 2

∴ r = -2a = -1 ⇒ α = -1,β = 2,γ = -4,δ = 8

(-1)^{2}-(-1)+P = 0 ⇒ p =-2

8,(-4)^{2}-4(-4)+q = 0 ⇒ q = -32

QUESTION: 21

Suppose a, b, c are in A.P. and a^{2}, b^{2}, c^{2} are in G.P. if a < b < c and a + b + c = 3/2, then the value of a is

[JEE 2002 ]

Solution:

2b = a + c & (b^{2})^{2} = a^{2}c^{2} & a < b < c

& a + b + c = 3/2 & b^{2} = __+__ ac

⇒ 3b = 3/2

roots of eqaution

Case - I

⇒ 4x^{2} - 4x + 1 = 0 = (2x- 1)^{2} = 0

⇒ a= c (but a < b < c) ∴ reject

Case -II

QUESTION: 22

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

Solution:

2 + 5 + 8 + .....+ T_{2n} = 57 + 59 + 61 + ....+ T_{n}

⇒ n [4 + 6n – 3] = n [57 + (n – 1)]

⇒ 6n + 1 = n + 56

⇒ 5n = 55

⇒ n = 11

QUESTION: 23

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

[JEE 2001, (Scr.)]

Solution:

QUESTION: 24

The first term of an infinite geometric progression is x and its sum is 5. Then

[JEE 2004 (Scr.)]

Solution:

QUESTION: 25

In the quadratic equati on ax^{2} + bx + c = 0, If Δ = b^{2} – 4ac and α + β, α^{2} + β^{2}, α^{3}+ β^{3}, are in G.P. where α , β are the roots of ax^{2} + bx + c = 0, then

[JEE 2005 (Scr.)]

Solution:

Δ = b2 — 4ac, quadratic ⇒ a ≠ 0

α + β, α^{2} + β^{2}, α^{3} + β^{3} in GP.

(α^{2} + β^{2})^{2 }= (α + β) (α^{3} + β^{3})

⇒ α^{4} + β^{4} + 2α^{2}β^{2 }= α^{4} + β^{4 } + αβ(α^{2} + β^{2})

⇒ αβ(α^{2} + β^{2} - 2αβ) = 0

⇒ αβ (α - β)^{2} = 0

⇒ cΔ = 0

QUESTION: 26

Let V_{r} d enote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r – 1). Let T_{r} = V_{ r + 1} – V_{r} – 2 and Q_{r} = T_{r + 1} – T_{r} for r = 1, 2,.....

The sum V_{1} + V_{2} + ....... + V_{n} is

Solution:

QUESTION: 27

Let V_{r} denote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r – 1). Let T_{r} = V_{ r + 1} – V_{r} – 2 and Q_{r} = T_{r + 1} – T_{r} for r = 1, 2,.....

T_{r} is always

Solution:

V_{r+1} - Vr

which is composite no.

QUESTION: 28

Le t V_{r} d enote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r – 1). Let T_{r} = V _{r + 1} – V_{r} – 2 and Q_{r} = T_{r + 1} – T_{r} for r = 1, 2,.....

Which one of the following is a correct statement ?

Solution:

T_{r }= 3r^{2} + 2r-1

T_{r+1} = 3(r+ 1)^{2} + 2(r + 1)-1

Q_{r} = T_{r+1} - T_{r} = 3(2r + 1)+ 2 (1)

Q_{r} = 6r + 5

Q_{r+1} = 6(r + 1)+ 5

common difference = Q_{r+ 1} - Q_{r} = 6

QUESTION: 29

Let A_{1}, G_{1}, H_{1} denote the arithmetic, geometric and harmoni c means, respectively, of two distinct positive numbers. For n ≥ 2, Let A_{n – 1} and H_{n – 1} have arithmetic, geometric and harmonic means as A_{n}, G_{n}, H_{n} respectively

Which one of the following statements is correct ?

Solution:

QUESTION: 30

G1,G2,…,Gn are said to be n geometric means between a and b if a,G1,…Gn,b is

Solution:

QUESTION: 31

Let A1, G1, H1 denote the arithmetic, geometric and harmonic means, respectively, of two distinct positive numbers. For n ≥ 2, let An−1 and Hn−1 has arithmetic, geometric and harmonic means as An, Gn, Hn respectively.

Which of the following statements is correct?

Solution:

A_{3} is A.M. of A_{1} an H_{1} and A_{1} > H_{1} ⇒ A_{1} > A_{2} > H_{1}

A_{3} is A.M. of A_{2} an H_{2} and A_{2} > H_{2} ⇒ A_{2} > A_{3} > H_{2}

∴ A_{1} > A_{2} > A_{3} > ...

QUESTION: 32

Let A_{1}, G_{1}, H_{1} denote the arithmetic, geometric and harmoni c means, respectively, of two distinct positive numbers. For n ≥ 2, Let A_{n – 1} and H_{n – 1} have arithmetic, geometric and harmonic means as A_{n}, G_{n}, H_{n} respectively

Solution:

As above A_{1} > H_{2} > H_{1} & A_{2} > H_{3} > H_{2}

∴ H_{1} > H_{2} < H_{3} < ......

QUESTION: 33

A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then

[JEE 2008, 4]

Solution:

(PS) × (ST) = (QS) × (SR)

......(i)

...(ii)

from (i) & (ii)

QUESTION: 34

Supose four distinct positive numbers a_{1}, a_{2}, a_{3}, a_{4} are i n G. P. Let b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, b_{3} = b_{2} + a_{3} and b_{4} = b_{3} + a_{4}.

Statement–1 : The numbers b_{1}, b_{2}, b_{3}, b_{4} are neither in A.P. nor in G.P.

Statement–2 : The numbers b_{1}, b_{2}, b_{3}, b_{4} are in H.P.

[JEE 2008]

Solution:

b_{1} = a_{1}

b_{2} = a_{1} + a_{2}

b_{3} = a_{1} + a_{2} + a_{3 }

b_{4 }= a_{1} + a_{2} + a_{3} + a_{4}

Hence b_{1}, b_{2}, b_{3}, b_{4} are neither in A.P. nor in G.P. nor in H.P.

QUESTION: 35

If the sum of first n terms of an A.P. is cn^{2}, then the sum of squares of these n terms is

[JEE 2009]

Solution:

QUESTION: 36

Let S_{k}, K = 1, 2, ...., 100 denote the sum of the infinite geometric series whose first term is and the common ratio is 1/k. Then the value of

Solution:

but k ≠ 1

QUESTION: 37

Let a_{1}, a_{2}, a_{3}, ....., a_{11} be real numbers satisfying a1 = 15, 27 – 2a_{2} > 0 and a_{k} = 2a_{k-1} – a_{k-2} for k = 3, 4, ...., 11

If then yhe value of is equal to

[JEE 2010]

Solution:

= a_{k} = 2a_{k-1} - a_{k-2}

⇒ a_{1}, a_{2}, a_{3} ...., a_{11} are in A.P. with common difference d.

a_{1} = 15,a_{2} = 15+d, a_{3} = 15+2d ..... a_{11} = 15+ 10d

⇒ 35d^{2} + 150d + 225 = 90

⇒ 35d^{2} + 150d + 135 = 0

⇒ 7d^{2} + 30d + 27 = 0

QUESTION: 38

The minim um value of the s um of re al numb ers a _{–5}, a_{–4}, 3a_{–3}, 1, a_{8 }and a_{10} with a > 0 is

[JEE 2011]

Solution:

Using AM __>__ GM

⇒ Sum = 8

QUESTION: 39

Let a_{1}, a_{2}, a_{3}, ..., a_{100} be an arithmetic progression with a_{1} = 3 and For any integer n with 1≤ n ≤ 20. let m=5n. If S_{m}/S_{n }does not depend on n, then a_{2} is

[JEE 2011]

Solution:

QUESTION: 40

Let a_{1}, a_{2}, a_{3}, ..... be in harmonic progression with a_{1} = 5 and a_{20} = 25. The least positive integer n for which a_{n} < 0 is

[JEE 2012]

Solution:

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