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What Are Phases

The phase is used to measure how far apart the waves are overlapping each other. What are the phases? Two waves are said to be “in phase” if the... View Article

What Is Ionization Air Purifier

An ionization air purifier will ionise the air, the dust particles get charged and are attracted to neutral surfaces like walls and floor. These... View Article

The value of [latex]0.16^{log_{2.5}} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}}+…\infty)[/latex] is equal to:

(1) 1 (2) 0 (3) -1 (4) None of these Solution: Here (⅓) + (1/32) + (1/33) …is a GP with a = ⅓ and r = ⅓ Use formula Sum = a/(1-r)... View Article

The First Two Terms Of A Geometric Progression Add Up To 12 The Sum Of The Third And The Fourth Terms Is 48 If Terms Of The Geometric Progression Are Alternately Positive And Negative Then The First Term Is

(1) 4 (2) - 4 (3) - 12 (4) 12 Solution: Let a , ar, ar2, ar3 be the 4 terms of GP. a+ar = 12 a(1+r) = 12 …(i) ar2+ar3 = 48... View Article

If 1 Sin X Sin2 X Upto Infinity Eq 4 2 Root 3 0 Less Than X Pi And X Not Equal Pi 2 Then X Is Equal To

(1) π/3 , 5π/6 (2) 2π/3, π/6 (3) π/3, 2π/3 (4) π/6, π/3 Solution: Given 1 + sin x + sin2 x+.. upto ∞ = 4 + 2√3 This is a GP with... View Article

The Value Of N For Which Xn 1 Yn 1 Xn Yn Is The Geometric Mean Of X And Y Is

(1) n = -½ (2) n = ½ (3) n = 1 (4) n = -1 Solution: Geometric mean of x and y = √xy ( xn+1 + yn+1 )/(xn+yn) = √xy ( xn+1 + yn+1... View Article

If Sum Of An Infinite Geometric Series Is 4 3 And First Term Is 3 4 Then Its Common Ratio Is

(1) 7/16 (2) 9/16 (3) 1/9 (4) 7/9 Solution: Given sum of infinite geometric series = a/(1-r) = 4/3 a = ¾ ¾(1-r) = 4/3 1-r =... View Article

The Sum Of First 20 Terms Of The Sequence 07 0 77 0 777 Is

(1) 7/81 (179 - 10-20) (2) 7/9 (99 - 10-20) (3) 7/81 (179 + 10-20) (4) 7/9 (99 + 10-20) Solution: S = 0.7 + 0.77 + 0.777 +... View Article

If The Second And Fifth Terms Of A Gp Are 24 And 3 Respectively Then The Sum Of The First Six Terms Is

(1) 181 (2) 181/2 (3) 189 (4) 189/2 Solution: Let a be the first term and r be the common ratio. So ar = 24 ..(i) ar4 = 3 ..(ii)... View Article

If 10 9 2 11 1 10 8 3 11 10 7 Equal K 10 9 Then K Is Equal To

(1) 100 (2) 110 (3) 121/100 (4) 441/100 Solution: Given (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 + ...+ 10(11)9 = k(10)9 Cancelling... View Article

The Sum Of N Terms Of Two Arithmetic Series Are In The Ratio 2n 3 6n 5 Then The Ratio Of Their 13th Terms Is

(1) 53 : 155 (2) 27 : 87 (3) 29 : 83 (4) 31 : 89 Solution: Let S1/S2= (2n + 3)/(6n + 5) S = (n/2)(2a+(n-1)d) Let a1, a2 be the first term... View Article

In An Arithmetic Progression The 24th Term Is 100 Then The Sum Of The First 47 Terms Of The Arithmetic Progression Is

(1) 2300 (2) 2350 (3) 2400 (4) 4700 Solution: Given a24 = a+23d = 100 Sum Sn = (n/2)(2a+(n-1)d) S47 = (47/2)(2a+46d = 47(a+23d)... View Article

Geometric Mean Of 7 7 Square 7 Cube 7 Power N Is

(1) 7(n+1)/2 (2) 7 (3) 7n/2 (4) 7n Solution: Let a1, a2...an be the n numbers in GP. Geometric mean = nth root of (a1. a2. ...an)... View Article

The Sum Of All Odd Numbers Between 1 And 1000 Which Are Divisible By 3 Is

(1) 83667 (2) 90000 (3) 83660 (4) None of these Solution: Here first term, a = 3 Second term = 9 Common difference, d = 6 Last... View Article

If Three Positive Real Numbers A B And C Are In Ap And Abc 4 Then The Minimum Possible Value Of B Is

(1) 23/2 (2) 22/3 (3) 21/3 (4) 25/2 Solution: Since a,b,c are in AP 2b = a+c Given abc = 4 AM ≥ GM (a+b+c)/3 ≥ (abc)⅓... View Article

If A1 A2 An Are In Ap With Common Difference D Not Equal 0 Then Sin D Sec A Is Equal To

(1) cot an + cot a1 (2) cot an - cot a1 (3) tan an - tan a1 (4) tan an - tan an-1 Solution: Given a1, a2….an are in AP.... View Article

If The First Second And Last Terms Of An Arithmetic Series Are A B And C Respectively Then The Number Of Terms Is

Solution: First term = a Second term = b Common difference, d = b-a Last term, (a+(n-1)d = c a+(n-1)(b-a) = c (n-1)(b-a) = (c-a) (n-1) =... View Article

If 5 9 13 N Terms 7 9 11 12 Terms 5 12 Then N Equals

(1) 5 (2) 6 (3) 9 (4) 12 Solution: Let S1 = 5+9+13+....n terms Here a = 5, d = 4 S1 = (n/2)(2a+(n-1)d) = (n/2)(10+(n-1)4) =... View Article

If The Sum Of First N Terms Of An Ap Is Cn2 Then The Sum Of Squares Of These N Terms Is

(1) n(4n2-1)c2/6 (2) n(4n2+1)c2/3 (3) n(4n2-1)c2/3 (4) none of these Solution: Given Sn = cn2 Sn-1 = c(n-1)2 = cn2+c-2cn Tn... View Article

An Ap Consists Of 23 Terms If The Sum Of The Three Terms In The Middle Is 141 And The Sum Of The Last Three Terms Is 261 Then The First Term Is

(1) 6 (2) 5 (3) 4 (4) 3 Solution: Let a1,a2,...a23 be the terms in AP. Let d be the common difference. Given a11+a12+a13 =... View Article