IIT JEE

Given
p A.P’s, each of which consists of n terms . If their first terms
are 1, 2, 3, -----, p and common differences are

1, 3, 5, ---, 2p –1 repectively , then sum of the terms of all the
progressions is .

(A) np(np+1)

(B) n(p+1)

(C) np(n+1)

(D) none of these .

The rth A. P. has first term r and common difference 2r-1. Hence sum of its n terms = .

The required sum =

= =

= . Hence (A) is the correct answer.

If log2, log(2^{x}-1) and
log(2^{x}+3) are in A.P. , then the value of x is .

(A) 5/2

(B) log_{2}5

(C) log_{3}5

(D) log_{5}3

2 log(2^{x}–1)
= log2 + log(2^{x }+3)

&⇒ ( 2^{x }–1)^{2} =
2. (2^{x} +3)

&⇒ (2^{x})^{2}
– 4.2^{x} – 5 = 0.

&⇒ ( 2^{x} – 5) ( 2^{x}
+1) = 0

&⇒ x = log_{2} 5 , as 2^{x}
+1 > 0

Hence (B) is the correct answer.

If a, b and c are distinct
positive real numbers and a^{2} +b^{2} +c^{2} =1,
then ab + bc +ca is

(A) less than 1

(B) equal to 1

(C) greater than 1

(D) any real number.

Since a and b are unequal , ( A.M. > G.M. for unequal numbers)

&⇒ a^{2} +b^{2}>
2ab

Similarly b^{2}
+c^{2} > 2bc and c^{2} +a^{2} > 2ca.

Hence 2(a^{2}
+b^{2} +c^{2} ) > 2(ab + bc +ca)

&⇒ ab +bc +ca < 1

Hence (A) is the correct answer.

If a, b and c are positive real numbers , then least value of (a+b+c) is

(A) 9

(B) 3

(C) 10/3

(D) none of these

Using A.M. ≥ G.M. , .

≥ (abc)^{1/3} and ≥

&⇒ .≥ 1

&⇒ ≥ 9 .

Equality will hold when a= b = c

Hence (A) is the correct answer. .

If first and (2n-1)th terms of an A.P. , G. P. and H.P. , are equal and their nth terms are a, b, c respectively , then .

(A) a+c = 2b

(B) a+c = b

(C) a ≥ b ≥ c

(D) ac –b^{2 }= 0

Let a be the first and b be the (2n-1)th term of an A.P. , G.P. and H.P. , then a, a, b will be in A.P. , a, b, b will be G.P. a, c, b will be in H.P. .

Hence a, b, c are respectively A. M. , G.M. and H.M. of a and b. Since A.M. ≥ G.M. ≥H.M. , a ≥ b ≥ c. .

Again a = , b^{2} = ab and c = .Hence ac-b^{2}
=0.

Hence (C) and (D) are correct answers.

Let p, q, r ∈
R^{+} and 27 pqr ≥ ( p + q + r)^{3} and 3p + 4q
+ 5r = 12 then

p^{3} + q^{4} + r^{5} is equal to

(A) 3

(B) 6

(C) 2

(D) none of these

27 pqr ≥
( p + q + r )^{3}

&⇒
( pqr)^{1/3} ≥

&⇒
p = q = r

Also 3 p + 4q + 5r =12

&⇒ p = q = r =1 .

Hence (A) is the correct answer.

If x_{i} >
0, i = 1, 2, . . . ., 50 and x_{1} +x_{2} + . . . + x_{50}
= 50, then the minimum value of equals
to

(A) 50

(B) (50)^{2}

(C) (50)^{3 }

(D) (50)^{4}

We have (x_{1} +x_{2} +x_{3}+
.. + x_{50}) ≥ (50)^{2}

[since A.M. ≥ H.M.].

&⇒
≥ 50.

Hence (A) is the correct answer.

The sum of first n terms of the
series is

equal to

(A) 2^{n} - n - 1

(B) 1-2^{-n}

(C) n + 2^{-n}-1

(D) 2^{n} -1

= n - = n +
2^{-n} –1 .

Hence (C) is the correct answer. .

If a, b, c are the p^{th}, q^{th}
r^{th} terms respectively of an H.P. then .

ab(p - q) + bc(q - r) + ca(r - p) equals to

(A) 1

(B) –1

(C) 0

(D) None of these

Let x be the first term and y be the c. d. of corresponding A.P. , then .

Multiplying (1) , (2) and (3) receptively by abc(q – r) , abc(r – p) , abc(p – q) and then adding we get

bc( q – r) + ca( r – p) + ab(p – q) = 0

Hence (C) is the correct answer. .

For 0 < q < p/2 , if

x = then

(A) xyz = xz + y

(B) xyz = xy +z

(C) xyz = x+y+z

(D) xyz = yz +x .

x = =
cosec^{2}q ,

y = sec^{2}
q , z =

&⇒
z
=

&⇒ xyz = xy + z . .

Also x + y = cosec^{2
}q + sec^{2 }q

=

= cosec^{2}q sec^{2}q = xy

&⇒ x + y + z = xy + z = xyz.

Hence (B), (C) are the correct answers. .