*Year-wise Solution of IIT-JEE paper on Sequences & Series*

**Q.[2004] An infinite G.P. has first term 'x' and sum '5'; then x belongs to**

(a) x < −10 (b) −10< x < 0 (c) 0< x <10 (d) x > 10

(a) x < −10 (b) −10< x < 0 (c) 0< x <10 (d) x > 10

Sol:- (c)

Let r be the common ratio then sum of infinite G.P. Series is

S = x/(1−r) ; − 1 < r < 1

So, 5 = x/(1−r) ;

⇒ r = (5−x)/5 ----------(1)

Now, − 1 < r < 1

[ Put the value of r from (1) ]

⇒ 0 < x < 10.

**Q.[2001] Let the positive numbers a,b,c,d be in A.P. Then abc,abd,acd,bcd are**

(a) not in A.P./G.P./H.P.

(b) in A.P.

(c) in G.P.

(d) in H.P.

(a) not in A.P./G.P./H.P.

(b) in A.P.

(c) in G.P.

(d) in H.P.

Sol:- (d)

Given a,b,c,d be in A.P.

A sequence in A.P. remains in A.P. , if each term is divided by a fixed non-zero number.

Now, dividing each term by abcd

So, 1/bcd,1/cda,1/abd,1/abc are in A.P.

So, bcd,cda,abd,abc are in H.P.

**Q.[2000] If a,b,c,d are positive real numbers such that a+b+c+d=2, then M=(a+b)(c+d) satisfies the relation**

(a) 0 ≤ M ≤ 1 (b) 1 ≤ M ≤ 2

(a) 0 ≤ M ≤ 1 (b) 1 ≤ M ≤ 2

**(c) 2 ≤ M ≤ 3 (d) 3 ≤ M ≤ 4**

Sol:- (a) Take two numbers a+b , c+d

Apply A.M. & G.M. inequality:

(a+b+c+d)² ≥ 4(a+b)(c+d)

⇒ (a+b)(c+d) ≤ 1

Since a,b,c,d are positive integers so, (a+b)(c+d) ≥ 0

Thus we have 0 ≤ M ≤ 1.