If The Pth Term Of A Gp Is X And Q Th Term Is Y Then The Nth Term Is
(1) (xn-p/yn-q)1/(p-q) (2) (xn+q/yn+p)1/(p-q) (3) (xn-q/yn-p)1/(p-q) (4) (xn-q/yn-q)1/(p+q) Solution: For a GP, pth term = arp-1 = x qth term... View Article
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Sum Of Infinity Of The Series 1 Plus 4 5 7 By 5 2 Plus 10 By 5 3 Is
(1) 7/16 (2) 5/16 (3) 104/64 (4) 35/16 Solution: Let S = 1+ ⅘ + 7/52 + 10/53 + …(i)... View Article
If A Is The Am Between A And B Then A 2b By A A Plus A 2a By A B
(1) -8 (2) 2 (3) 4 (4) -4 Solution: Since A is the AM of a and b, A =... View Article
In An Ap Sm Sn Eq M2 N2 The Ratio Of P2 Th Term To Q2 Term Is
(1) (2p2+1)/(2q2+1) (2) (2p2-1)/(2q2-1) (3) (2p-1)/(2q-1) (4) (p2-2)/(q2-2) Solution: For an AP, S = (n/2)(2a+(n-1)d) Sm : Sn = m2... View Article
If 1 By A 1 By H 1 By B Are In Ap Then H Plus A By H A Plus H Plus B By H B
(1) 2 (2) 4 (3) 0 (4) 1 Solution: Given 1/a, 1/H, 1/b are in AP. 2/H = 1/a +... View Article
If A1 A2 A10 Be In Ap 1 By H1 1 By H2 1 By H10 Be In Ap A1 Eq H1 Eq 2 A10 Eq H10 Eq 3 Then A4h7
(1) 1/6 (2) 6 (3) 3 (4) 2 Solution: Given a1, a2, …a10 be in AP. a1 = h1 =... View Article
If A 4 B Are In Ap And A 2 B Are In Gp Then 1 By A 1 1 By B Are In
(1) AP (2) GP (3) HP (4) None of these Solution: Since a, 4, b are in AP 8 =... View Article
If For An Ap An A1 A5 A15 A26 A36 A40 Eq 210 Then S40 Eq
(1) 2100 (2) 700 (3) 1400 (4) None of these Solution: a1 + a5 + a15 + a26 + a36... View Article
If The Function F Satisfies The Relation F X Plus Y Eq Fx Fy For All X Y Belongs To N Further If F1 Eq 3
(1) 4 (2) 2 (3) 1 (4) 3 Solution: Given f (x+y) = f(x) f(y) Also f(1) = 3 ∑x... View Article
If 1 Logy X Logz Y 15 Logx Z Are In Ap Then The Common Difference Of This Ap Is
(1) 1 (2) 2 (3) -2 (4) 3 Solution: Given 1, logy x, logz y, -15 logx z are in... View Article
The Sum Of 10 Terms Of The Series 12 16 24 40 Is
(1) 4094 (2) 4296 (3) 4172 (4) none of these Solution: Let S = 12 + 16 + 24 +... View Article
The Coefficient Of X 8 In The Product X 1 X 2 X 3x 10 Is
(1) 1024 (2) 1300 (3) 1320 (4) 1360 Solution: We know (x-a)(x-b) = x2-(a+b)x+ab (x-1)(x-2) = x2-(2+1)x + 2×1 (x-a1)(x-a2)…(x-an)... View Article
A Sample Of 4 Items Is Drawn At Random Without Replacement From A Lot Of 10 Items Containing 3 Defective If X Denotes The Number Of Defective
1) 3/10 2) 4/5 3) 1/2 4) 1/6 Solution: Option (2) 4/5 Given that, the items are chosen without replacement.... View Article
If X Has A Binomial Distribution With Mean Np And Variance Npq Then Pxk Pxk1 Is
1) [(n-k)/(k-1)].[p/q] 2) [(n-k+1)/(k)].[p/q] 3) [(n+1)/k].[p/q] 4) [(n-1)/(k+1)].[p/q] Solution: Option (2) [(n-k+1)/(k) Given that, mean = np, variance= npq. P(X=k)... View Article
The Mean And Variance Of A Binomial Distribution Are 4 And 2 Respectively Then The Probability Of 2 Successes Is
1) 28/256 2) 219/256 3) 128/256 4) 37/256 Solution: Option (1) 28/256 Given: Mean = np = 4 Variance, npq=... View Article
A Dice Is Thrown Two Times If Getting The Odd Number Is Considered As Success Then The Probability Of Two Successes Is
1) 1/2 2) 3/4 3) 2/3 4) 1/4 Solution: Option (4) ¼ p=3/6 = ½ q= ½ n= 2, and... View Article
A Coin Is Tossed 3 Times The Probability Of Obtaining At Least Two Heads Is
1) 1/8 2) 3/8 3) 1/2 4) 2/3 Solution: Option (3) ½ p= ½, q= ½, n= 3 By using... View Article
If A Dice Is Thrown Twice The Probability Of Occurrence Of 4 At Least Once Is
1) 11/36 2) 7/12 3) 35/36 4) None of these Solution: Option (1) 11/36 p= ⅙, q= ⅚, n= 2... View Article
A Coin Is Tossed 10 Times The Probability Of Getting Exactly Six Heads Is
1) 512/513 2) 105/512 3) 100/153 4) 10C6 Solution: Option (2) 105/512 Let x be the number of heads that... View Article
In A Binomial Distribution The Probability Of Getting A Success Is 14 And Standard Deviation Is 3 Then Its Mean Is
1) 6 2) 8 3) 12 4) 10 Solution: Option (3) 12 Given: p= ¼, standard deviation deviation = 3... View Article
