When 1 A 1 C 1 A B 1 C B 0 And B Not Eq A C Then A B C Are In (1) AP (2) GP (3) HP (4) None of these Solution: Given 1/a + 1/c + 1/(a-b) + 1/(c-b) = 0 Rearranging the terms [(1/a) + 1/(c-b)]... View Article
If A Bx A Bx B Cxb Cx C Dx C Dx X 0 Then A B C D Are In (1) AP (2) GP (3) HP (4) None of these Solution: Given (a+bx)/(a-bx) = (b+cx)/(b-cx) = (c+dx)/(c-dx) (a+bx)/(a-bx) =... View Article
If X Plus Y By 2 Y Y Plus Z By 2 Are In Hp Then X Y Z Are In (1) AP (2) GP (3) HP (4) None of these Solution: Given (x+y)/2, y, (y+z)/2 are in HP. So 2/(x+y), 1/y, 2/(y+z) are in AP. (1/y)... View Article
If 9 Ams And Hms Are Inserted Between 2 And 3 And If The Harmonic Mean H Is Corresponding To Arithmetic Mean A Then A Plus 6 By H Is Equal To (1) 1 (2) 3 (3) 5 (4) 6 Solution: Let A1,A2 ...A9 are the AM’s and H1, H2, ...H9 are the HM’s between a and b. 9 AM’s and HM’s... View Article
If A1 A2 Are Two Ams Between Two Numbers A And B And G1 G2 Be Two Gms Between Same Two Numbers Then A1 Plus A2 By G1g2 (1) (a+b)/ab (2) (a+b)/2ab (3) 2ab/(a+b) (4) ab/(a+b) Solution: Let a, A1, A2, b are in AP. Then (a+b)/2 = (A1+A2)/2 (A1+A2) =... View Article
If A And B Two Different Positive Real Numbers Then Which Of The Following Relation Is True (1) 2√(ab) > (a+b) (2) 2√(ab) < (a+b) (3) 2√(ab) = (a+b) (4) none of these Solution: We know AM ≥ GM (a+b)/2 ≥ √(ab) (a+b)... View Article
The Value Of 1 By 1 Plus A By 2 Plus A Plus 1 By 2 Plus A By 3 Plus A Infinity Is Where A Is A Constant (1) 1/(1+a) (2) 2/(1+a) (3) ∞ (4) None of these Solution: 1/(1+a)(2+a) + 1/(2+a)(3+a) + 1/(3+a)(4+a) +..... = 1/(1+a) - 1/(2+a) +... View Article
If 1 By 1 4 Plus 1 By 2 4 Plus 1 By 3 4 Infinity Pi 4 90 Then The Value Of 1 By 1 4 Plus 1 By 3 4 Plus 1 By 5 4 Infinity Is (1) π4/96 (2) π4/45 (3) 89 π4/90 (4) none of these Solution: Given 1/14 + 1/24 + 1/34 + 1/44 + ...∞ = π4/90 (1/14 + 1/34 + 1/54... View Article
If Tn Eq 1 By 4 N Plus 2 N Plus 3 For N Eq 1 2 3 Then 1 T1 1 T2 1 T3 1 T2003 (1) 4006/3006 (2) 4003/3007 (3) 4006/3008 (4) 4006/3009 Solution: Given tn = (1/4)(n+2)(n+3) t1 = (1/4)(3×4) t2 = (¼)×4×5 t3 =... View Article
Four Numbers Are In Arithmetic Progression The Sum Of The First And Last Term Is 8 And The Product Of Both Middle Terms Is 15 The Least Number Of The Series Is (1) 4 (2) 3 (3) 2 (4) 1 Solution: Let a-3d, a-d , a+d, a+3d are in AP. The sum of the first and last term is 8. a-3d+a+3d = 8... View Article
If A B C D E Are In Ap Then The Value Of A B 4c 4d E In Terms Of A If Possible Is (1) 4a (2) 2a (3) 3 (4) None of these Solution: Given a,b,c,d,e are in A.P. Let D be the common difference of AP. b = a+D c =... View Article
If 1 Log9 3 1 X Plus2 Log3 4 3x 1 Are In Ap Then X Equals (1) log3 4 (2) 1 – log3 4 (3) 1 – log4 3 (4) log4 3 Solution: Given 1, log9 (31– x + 2), log3(4.3x-1) are in A.P So 2log9 (31– x... View Article
Let Tr Be The Rth Term Of An Ap For R 1 2 3 If For Some Positive Integers M N We Have Tm Eq 1 By N And Tn Eq 1 By M Then Tmn Equal (1) 1/mn (2) 1/m + 1/n (3) 1 (4) 0 Solution: Let a be the first term and d be the common difference of AP. rth term = Tr Tm... View Article
If Am Denotes The Mth Term Of An Ap Then Am (1) 2/(am+k + am-k) (2) (am+k - am-k)/2 (3) (am+k + am-k)/2 (4) None of these Solution: Let a be the first term and d be the... View Article
If The Pth Qth And Rth Term Of An Arithmetic Sequence Are A B And C Respectively Then The Value Of (1) 1 (2) –1 (3) 0 (4) 1/2 Solution: Let A be the first term and D be the common difference. pth term = A+(p-1)D = a a = A-D+pD... View Article
The 9th Term Of The Series 27 Plus 9 Plus 5 And 2 By 5 Plus 3 And 6 By 7 Will Be (1) (2) 10/17 (3) 16/27 (4) 17/27 Solution: = 27 + 27/3 + 27/5 + 27/7 +....27/(2n-1) Tn = 27/(2n-1) T9 = 27/(18-1) = 27/17... View Article
The First And Last Terms Of A Gp Are A And L Respectively R Being Its Common Ratio Then The Number Of Terms In This Gp Is (1) (log l - log a)/ log r (2) 1 - (log l - log a)/ log r (3) (log a - log l)/ log r (4) 1 + (log l - log a)/ log r Solution:... View Article
Pth Term Of The Series 3 1 By N Plus 3 2 By N Plus 3 3 By N Will Be (1) 3 + p/n (2) 3 - p/n (3) 3 + n/p (4) 3 - n/p Solution: Here first term, a = 3 - 1/n Common difference, d = -1/n Pth term, ap... View Article
If Arithmetic Mean Of Two Positive Numbers Is A Their Geometric Mean Is G And Harmonic Mean Is H Then H Is Equal To (1) G2/A (2) G/A2 (3) A2/G2 (4) A/G2 Solution: Let a and b be two numbers. AM, A = (a+b)/2 GM, G = √(ab) HM, H = 2ab/(a+b) =... View Article
If A1 A2 An Are Positive Real Numbers Whose Product Is A Fixed Number C Then The Minimum Value Of A1 A2 An 1 2an Is (1) n(2c)1/n (2) (n + 1)c1/n (3) 2nc1/n (4) (n + 1)(2c)1/n Solution: Given product, a1a2...an = c a1a2…(an-1)(2an) = 2c ….(i) We... View Article