If Ax Equals By Equals Cz And A B C Are In Gp Then X Y Z Are In (1) AP (2) GP (3) HP (4) None of these Solution: Let ax = by = cz = p a = p1/x b = p1/y c = p1/z b2 = ac (since a,b,c... View Article
Log3 2 Log6 2 Log12 2 Are In (1) A.P (2) G.P (3) H.P (4) None of these Solution: Let x = log3 2 y = log6 2 z = log12 2 1/x = log2 3 1/y = log2 6 = log2 (2×3)... View Article
The Harmonic Mean Between Two Numbers Is 14 And 2 By 5 And The Geometric Mean Is 24 The Greater Number Between Them Is (1) 72 (2) 54 (3) 36 (4) none of these Solution: Let a and b be the numbers, GM = √(ab) = 24 ab = 576 a = 576/b …(i) HM =... View Article
If X A Equals X B By 2 Zb By 2 Equals Zc Then A B C Are In (1) A.P (2) G.P (3) H.P (4) None of these Solution: Given xa = xb/2 zb/2 = zc xa = (xz)b/2 zc = (xz)b/2 log xa = log... View Article
If Am Of Two Terms Is 9 And Hm Is 36 Then Gm Will Be (1) 18 (2) 12 (3) 16 (4) None of these Solution: Given AM = 9 HM = 36 We know GM2 = AM (HM) GM2 = 9×36 GM = 3×6 = 18 Hence... View Article
The Geometric Mean Of Two Numbers Is 6 And Their Arithmetic Mean Is 6 5 The Numbers Are (1) (3, 12) (2) (4, 9) (3) (2, 18) (4) (7, 6) Solution: Let a and b be the numbers. GM = 6 √(ab) = 6 ab = 36 b = 36/a …(i) AM... View Article
If Logx Y Logz X Logy Z Are In Gp Xyz 64 And X3 Y3 Z3 Are In Ap Then (1) x = y = z (2) x = 4 (3) x, y, z are in G.P (4) All of these Solution: Since logx y, logz x, logy z are in G.P (logz x)2= logx... View Article
If 2y A Is The Hm Between Y X And Y Z Then X A Y A Z A Are In (1) A.P (2) G.P (3) H.P (4) None of these Solution: Given 2(y - a) is the H.M between y - x and y - z. So (y - x), 2(y - a), (y -... View Article
If A Plus B By 1 Ab B B Plus C By 1 Bc Are In Ap Then A 1 By B C Are In (1) A.P (2) G.P (3) H.P (4) None of these Solution: Since (a+b)/(1-ab), b, (b+c)/(1-bc) are in AP 2b = [(a+b)/(1-ab)] +... View Article
If A B C Be In Gp And A Plus X B Plus X C Plus X In Hp Then The Value Of X Is A B C Are Distinct Numbers (1) c (2) b (3) a (4) None of these Solution: Since a, b, c are in GP b2 = ac …(i) a + x, b + x, c + x, in H.P So 1/(a+x) ,... View Article
An Ap A Gp And A Hp Have The Same The First And Last Terms And The Same Odd Number Of Terms The Middle Terms Of The Three Series Are In (1) A.P (2) G.P (3) H.P (4) None of these Solution: Let a and b be the first and last terms of AP, GP, and HP and they have the... View Article
If The Ratio Of Two Numbers Be 9 Is To 1 Then The Ratio Of Geometric And Harmonic Means Between Them Will Be (1) 1 : 9 (2) 5 : 3 (3) 3 : 5 (4) 2 : 5 Solution: Let a and b be the numbers. a/b = 9/1 a = 9b …(i) GM = √(ab) HM = 2ab/(a+b)... View Article
The Numbers Root 2 Plus 1 1 Root 2 1 Will Be In (1) A.P (2) G.P (3) H.P (4) None of these Solution: Given numbers are √2 +1, 1, √2 -1 Check the condition for AP 2b = a+c 2b = 1... View Article
In The Four Numbers First Three Are In Gp And Last Three Are In Ap Whose Common Difference Is 6 If The First And Last Numbers Are Same Then First Will Be (1) 2 (2) 4 (3) 6 (4) 8 Solution: Let a-d, a, a+d be the numbers in AP. Given the first number is the same as the fourth number.... View Article
If A B C Are In Ap And A2 B2 C2 Are In Hp Then (1) a = b = c (2) 2b = 3a + c (3) b2 = √(ac/8) (4) None of these Solution: Given that a, b, c are in A. P. 2b = a + c ……. (i)... View Article
If Gm Equals 18 And Am Equals 27 Then Hm Is (1) 1/18 (2) 1/12 (3) 12 (4) 9√6 Solution: We know GM2 = AM×HM AM = 27 GM = 18 182 = 27 HM HM = 182/27 = 324/27 = 12 Hence... View Article
If A B C Are In Ap Then 3a 3b 3c Shall Be In (1) A.P (2) G.P (3) H.P (4) None of these Solution: Since a, b and c are in A.P then b - a = c - b 3(b-a) = 3(c-b) 3b/3a =... View Article
Given Ax Eq By Eq Cz Eq Du And A B C D Are In Gp Then X Y Z U Are In (1) A.P (2) G.P (3) H.P (4) None of these Solution: Let ax = by = cz = du = k So a = k1/x …(i) b = k1/y …(ii) c = k1/z …(iii)... View Article
If A B C Are In Gp A B C A B C Are In Hp Then A Plus 4b Plus C Is Equal To (1) 0 (2) 1 (3) -1 (4) None of these Solution: Since a, b, c are in GP b2 = ac …(i) Since a - b, c - a, b - c are in H.P 2/(c-a) =... View Article
If A B C Are In Ap B C D Are In Gp And C D E Are In Hp Then A C E Are In (1) No particular order (2) A.P (3) G.P (4) H.P Solution: Since a, b, c are in AP 2b = a+c …(i) Since b, c, d are in GP c2 = bd... View Article