The Greatest Value Of N For Which 1 Plus 1 By 2 Plus 1 By 2 2plus 1 By 2 N Less Than 2 Is
The greatest value of n for which 1 + 1/2 + 1/22 +…+1/2n < 2 is (n ∈N) (1) 100... View Article
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If Sec X Y Sec X And Sec Xplusy Are In Ap Then Cos X Sec Y By 2 Equal
(1) ±√2 (2) ±1/√2 (3) ±2 (4) ±1/2 Solution: Given sec (x-y), sec x and sec (x+y) are in AP.... View Article
1 By 2 Plus 3 By 4 Plus 7 By 8 Plus 15 Y 16 Plus N Terms Equal
1/2 + 3/4 + 7/8 + 15/16 +…n terms = (1) n+2-n-1 (2) 2-n-n +1 (3) (2-n-n +1)/4 (4) 2-n... View Article
If 1 By A 1 By B 1 By C Are In Ap Then 1 By A Plus 1 By B 1 By C 1 By B Plus 1 By C 1 By A Equal
(1) (4b2-3ac)/abc (2) 4/ac – 3/b2 (3) 4/ac – 5/b2 (4) (4b2+ 3ac)/ab2c Solution: Given 1/a, 1/b, 1/c are in... View Article
The Sum Of The Series 3 By 4 Plus 5 By 36 Plus 7 By 144 Upto 11 Terms Is
(1) 120/121 (2) 143/144 (3) 1 (4) 144/143 Solution: Let S = 3/4 + 5/36 + 7/144 +…upto 11 terms... View Article
6th Term Of The Sequence 7 By 3 35 By 6 121 By 12 335 By 24 Is
(1) 2113/96 (2) 2112/96 (3) 2111/96 (4) 865/48 Solution: Given sequence 7/3, 35/6, 121/12, 335/24,… = 2+1/3 , 6-1/6, 10+1/12,... View Article
If The Sides Of A Triangle Abc Are In Ap And The Greatest Angle Is Double The Smallest The Ratio Of The Sides Of Abc Is
(1) 3 : 4 : 5 (2) 5 : 12 : 13 (3) 4 : 5 : 6 (4) 5... View Article
1 Plus 4 By 5 Plus 7 By 52 Plus 10 By 53 Upto Infinity Equal
(1) 16/35 (2) 11/8 (3) 35/16 (4) 7/16 Solution: Let S = 1+ 4/5 + 7/52 + 10/53 +…upto ∞... View Article
If The Angles A B C Of A Triangle Abc Are In Ap Then
1) c2 = a2 + b2 – ab2 2) c2 = a2 + b2 3) b2 = a2 + c2... View Article
If The Numbers P Q R Are In Ap Then 2 P2 2 Pq 2 Pr Are In
(1) A.P. (2) G.P. (3) H.P. (4) A.G.P. Solution: Given p,q,r are in AP. 2q = p+r (2p)2q = (2p)p+r... View Article
3 By 1 2 Plus 5 By 1 2 Plus 2 2 Plus 7 By 1 2 Plus 2 2 Plus 32 N Terms Equal
(1) 6n2/(n+1) (2) 6n/(n+1) (3) 6(2n-1)/(n+1) (4) 3(n2+1)/(n+1) Solution: Given series 3/12 + 5/12+22 + 7/(12+22+32) +…n terms Hete tn... View Article
The Nth Term Of The Sequence 1 By 1 Plus Root X 1 By 1 X 1 By 1 Root X Is
(1) (1+√x(n2-2))/(1-x) (2) (1+√x(n-1))/(1+√x) (3) (1+√x(n-2))/(1-x) (4) (3-√x(n+2))/3(1-x) Solution: Given series 1/1+√x, 1/1-x, 1/1-√x,… Here d = (1/1-x) – (1/1+√x)... View Article
If The Sides Of A Right Triangle Are In Ap Then The Sum Of The Sines Of The Two Acute Angles Is
(1) 7/5 (2) 8/5 (3) 1/5 (4) 6/5 Solution: Let the sides of the triangle be a-d, a, a+d. (a+d)2... View Article
The Numbers Of Terms In The Ap A B C X Is
(1) (x+b+a)/(c-a) (2) (x+b-2a)/(c-b) (3) (x+b+2a)/(c-b) (4) (x-b+a)/(c-b) Solution: Given a,b,c…x are in AP Common difference = b-a = c-b... View Article
The Sum Of The Series A A Plus D Plus A Plus 2d Up To 50 Terms Is
(1) -50d (2) 25d (3) a + 50d (4) -25d Solution: a – (a + d) + (a + 2d)... View Article
Sum Of Products Of First N Natural Numbers Taken Two At A Time Is
(1) n(n2-1)(3n+2)/24 (2) n(n+1)2(3n+2)/74 (3) n2(n+1)(3n+2)/48 (4) n(n+1)(n+2)(3n+2)/96 Solution: We know (x1 + x2 + …xn)2 = x12 +x22 +….xn2... View Article
If The First Second And Last Terms Of An Ap Are A B And 2a Respectively The Sum Of The Series Is
(1) ab/2(b-a) (2) 3ab/2(b-a) (3) ab/(b-a) (4) none of these Solution: Given first term = a Second term = b... View Article
If An Is An Ap Then A12 A22 A32 A42 A992 A1002 Equal
(1) (50/99)(a12-a1002) (2) (100/99)(a1002-a12) (3) (50/51)(a12-a1002) (4) none of these Solution: Given a1, a2, a3….an is in AP. a2-a1 =... View Article
If The Sum Of The Roots Of The Equation Ax2 Plus Bx Plus C Equal 0 Is Equal To The Sum Of The Squares Of Their Reciprocals Then Bc2 Ca2 Ab2 Are In
(1) AP (2) GP (3) HP (4) AGP Solution: Let p and q be the roots of the equation ax2... View Article
If Sn Equal An Plus Bn2 For An Ap Where A And B Are Constants Then Common Difference Of Ap Will Be
(1) 2b (2) a + b (3) 2a (4) a – b Solution: Given Sn = an+bn2 S1 = a+b... View Article
