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### IIT-JEE Previous Year Solved Paper On SEQUENCES & SERIES

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

Q. 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

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. 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.

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. 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

(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.