Q 1 :

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]

  • 22

     

  • 23

     

  • 24

     

  • 25

     

(4)

∵  a1,a2,a3,… are in H.P.

∴ 1a1,1a2,1a3,… are in A.P.

∴ 1a1=15  and  1a20=125

1a1+19d=1a20 ⇒ 15+19d=125 ⇒ d=-4475

Now 1an=15+(n-1)(-4475)

Clearly an<0, if 1an<0 ⇒ 15-4n475+4475<0

⇒-4n<-99  or  n>994=2434                   ∴  n≥25

∴ Least value of n is 25.



Q 2 :

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

  • NOT in A.P./G.P./H.P.

     

  • in A.P.

     

  • in G.P.

     

  • in H.P.

     

(4)

a,b,c,d are in A.P.   ∴ d,c,b,a are also in A.P.

⇒dabcd, cabcd, babcd, aabcd are also in A.P.

⇒1abc, 1abd, 1acd, 1bcd are in A.P.

∴ abc, abd, acd, bcd are in H.P.



Q 3 :

Let m be the minimum possible value of log3(3y1+3y2+3y3), where y1,y2,y3 are real numbers for which y1+y2+y3=9. Let M be the maximum possible value of (log3x1+log3x2+log3x3), where x1,x2,x3 are positive real numbers for which x1+x2+x3=9. Then the value of log2(m3)+log3(M2) is _____ .        [2020]



(8)

By AM--GM inequality

AM≥GM

∴ 3y1+3y2+3y33≥[3(y1+y2+y3)]13

⇒3y1+3y2+3y3≥34

⇒log3(3y1+3y2+3y3)≥4⇒m=4

∵ log3x1+log3x2+log3x3=log3(x1x2x3)

Again by AM-GM inequality

AM≥GM

x1+x2+x33≥x1x2x33⇒x1x2x3≤27

⇒log3(x1x2x3)≤log3(33)

⇒log3x1+log3x2+log3x3≤3⇒M=3

Now, log2(m3)+log3(M2)=6+2=8



Q 4 :

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                          [2008]

  • 1PS+1ST<2QS×SR

     

  • 1PS+1ST>2QS×SR

     

  • 1PS+1ST<4QR

     

  • 1PS+1ST>4QR

     

Select one or more options

(2, 4)

We know by geometry PS×ST=QS×SR            ...(i)

∵ S is not the centre of circumcircle, PS≠ST

And we know that for two unequal real numbers, H.M. < G.M.

⇒21PS+1ST<PS×ST⇒1PS+1ST>2PS×ST

⇒1PS+1ST>2QS×SR  [using eqn (i)] ...(ii)

∴ (2) is the correct option.

Also QS×SR<QS+SR2         (∵GM<AM)

⇒1QS×SR>2QR⇒2QS×SR>4QR            ...(iii)

From equations (ii) and (iii), 1PS+1ST>4QR

∴ (4) is also the correct option.



Q 5 :

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 have arithmetic, geometric and harmonic means as An,Gn,Hn respectively.                                [2007]

Q.   Which one of the following statements is correct?

  • G1>G2>G3>⋯

     

  • G1<G2<G3<⋯

     

  • G1=G2=G3=⋯

     

  • G1<G3<G5<⋯ and G2>G4>G6>⋯

     

(3)

Given A1=a+b2,  G1=ab,  H1=2aba+b

Also An=An-1+Hn-12,  Gn=An-1Hn-1

Hn=2An-1Hn-1An-1+Hn-1

⇒Gn2=AnHn⇒AnHn=An-1Hn-1

Similarly we can prove

     AnHn=An-1Hn-1=An-2Hn-2=⋯=A1H1

⇒AnHn=ab

∴ G12=G22=G32=⋯=ab

⇒G1=G2=G3=⋯=ab



Q 6 :

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 have arithmetic, geometric and harmonic means as An,Gn,Hn respectively.                    [2007]

 

Q.   Which one of the following statements is correct?

  • A1>A2>A3>...

     

  • A1<A2<A3<...

     

  • A1>A3>A5>... and  A2<A4<A6<...

     

  • A1<A3<A5<... and  A2>A4>A6>...

     

(1)

An=An-1+Hn-12

∴  An-An-1=An-1+Hn-12-An-1

                              =Hn-1-An-12<0  (∵An-1>Hn-1)

⇒  An<An-1  or  An-1>An

∴  A1>A2>A3>...



Q 7 :

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 have arithmetic, geometric and harmonic means as An,Gn,Hn respectively.               [2007]

 

Q.  Which one of the following statements is correct?

  • H1>H2>H3>...

     

  • H1<H2<H3<...

     

  • H1>H3>H5>... and  H2<H4<H6<...

     

  • H1<H3<H5<... and H2>H4>H6>...

     

(2)

AnHn=ab ⇒ Hn=abAn

∵  1An-1<1An ⇒ Hn-1<Hn

∴  H1<H2<H3<....



Q 8 :

Suppose four distinct positive numbers a1,a2,a3,a4 are in G.P. Let b1=a1, b2=b1+a2, b3=b2+a3, and b4=b3+a4.

STATEMENT-1: The numbers b1,b2,b3,b4 are neither in A.P. nor in G.P.

STATEMENT-2: The numbers b1,b2,b3,b4 are in H.P.                            [2008]

  • STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1

     

  • STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1

     

  • STATEMENT - 1 is True, STATEMENT - 2 is False

     

  • STATEMENT - 1 is False, STATEMENT - 2 is True

     

(3)

Given: a1,a2,a3,a4 are in G.P.

Then b1,b2,b3,b4 are the numbers  a1, a1+a2, a1+a2+a3, a1+a2+a3+a4

or   a, a+ar, a+ar+ar2, a+ar+ar2+ar3

Since, above numbers are neither in A.P. nor in G.P. Therefore,

statement 1 is true.

Also   1a, 1a+ar, 1a+ar+ar2, 1a+ar+ar2+ar3 are not in A.P.

∴  b1,b2,b3,b4 are not in H.P.

∴  Statement 2 is false.