If The Sum Of N Terms Of A Gp Is 255 And Nth Terms Is 128 And Common Ratio Is 2 Then First Term Will Be (1) 1 (2) 3 (3) 7 (4) None of these Solution: Given r = 2 an = 128 Sn = 255 an = arn-1 a×2n-1 = 128 Multiply both sides by... View Article
The Sum Of First Two Terms Of A Gp Is 1 And Every Term Of This Series Is Twice Of Its Previous Term Then The First Term Will Be (1) 1/4 (2) 1/3 (3) 2/3 (4) 3/4 Solution: Let a, ar, ar2 … be in GP a+ar = 1 a(1+r) = 1 ..(i) Also given every term of this... View Article
If The Sum Of An Infinite Gp Be 9 And The Sum Of First Two Terms Be 5 Then The Common Ratio Is (1) 1/3 (2) 3/2 (3) 3/4 (4) 2/3 Solution: Let a be first term and r be the common ratio of a GP. Sum of infinite GP = a/(1-r) = 9... View Article
The Third Term Of A Gp Is The Square Of First Term If The Second Term Is 8 Then The 6th Term Is (1) 120 (2) 124 (3) 128 (4) 132 Solution: Let a, ar, ar2 be the GP. Given ar2 = a2 So a = r2 Also ar = 8 So r3 = 8 r = 2 a = r2 = 4... View Article
If The Roots Of The Cubic Equation Ax3 Plus Bx2 Plus Cx Plus D Eq 0 Are In Gp Then (1) c3a = b3d (2) ca3 = bd3 (3) a3b = c3d (4) ab3 = cd3 Solution: Let α, β, γ are the roots of the equation ax3 + bx2 + cx + d = 0... View Article
The Value Of I I2 Plus I3 I4 I100 Equal Is (1) i (2) -i (3) 0 (4) 1 + i Solution: Given series i - i2 + i3 - i4 + ... - i100 Here a = i Common ratio, r = -i Sum of n... View Article
The 20th Term Of The Series 2 4 Plus 4 6 Plus 6 8 Will Be (1) 1600 (2) 1680 (3) 420 (4) 840 Solution: Given series 2 × 4 + 4 × 6 + 6 × 8 +.... Tn = 2n(2n+2) = 4n2+4n T20 = 4(20)2+4×20... View Article
If The Third Term Of A Gp Is 4 Then The Product Of Its First 5 Terms Is (1) 43 (2) 44 (3) 45 (4) None of these Solution: Given third term of a GP, ar2 = 4 Product of first 5 terms =... View Article
If The Ratio Of The Sum Of First Three Terms And The Sum Of First Six Terms Of A Gp Be 125 152 Then The Common Ratio R Is (1) 3/5 (2) 5/3 (3) 2/3 (4) 3/2 Solution: Sum of n terms of GP, Sn = a(rn-1)/(r-1) S3 = a(r3-1)/(r-1) S6 = a(r6-1)/(r-1) Given... View Article
If X 2x Plus 2 3x Plus 3 Are In Gp Then The Fourth Term Is (1) 27 (2) -27 (3) 13.5 (4) -13.5 Solution: Since x, 2x + 2, 3x + 3, are in G.P (2x+2)2 = x(3x+3) 4x2+8x+4 = 3x2+3x x2 + 5x + 4... View Article
The 5th Term Of The Series 10 By 9 1 By 3 Root 20 By 3 Is (1) 1/3 (2) 1 (3) 2/5 (4) √2/3 Solution: Given series 10/9, (⅓)√(20/3), 2/3,.... Check whether the given series is GP. Common... View Article
The Number Which Should Be Added To The Numbers 2 14 62 So That The Resulting Numbers May Be In Gp Is (1) 1 (2) 2 (3) 3 (4) 4 Solution: Let x be the number added to 2, 14, 62 so that the resulting numbers may be in GP. So (2+x),... View Article
If The First Term Of A Gp Be 5 And Common Ratio Be 5 Then Which Term Is 3125 (1) 6th (2) 5th (3) 7th (4) 8th Solution: Given a = 5 Common ratio, r = -5 nth term of GP = arn-1 3125 = 5(-5)n-1 (-5)n-1... View Article
If A B C Are In Gp Then (1) a(b2 + a2) = c(b2+c2) (2) a(b2 + c2) = c(a2 + b2) (3) a2(b+c) = c2 (a+b) (4) None of these Solution: Since a, b, c are in GP... View Article
1 Plus 5 Plus 14 Plus 30 Plus N Terms (1) (n+2)(n+3)/12 (2) n(n+1)(n+5)/12 (3) n(n+2)(n+3)/12 (4) n(n+1)2(n+2)/12 Solution: Let Sn = 1 + 5 + 14 + 30 + ... n terms = 12... View Article
The Number 1111 91 Times Is A (1) Even number (2) Prime number (3) Not prime (4) None of these Solution: Let S = 1+10+102+...1090 (since 91 terms) Here a = 1,... View Article
If The Sum Of First 6 Terms Is 9 Times To The Sum Of First 3 Terms Of The Same Gp Then Common Ratio Of The Series Will Be (1) -2 (2) 2 (3) 1 (4) 1/2 Solution: Let a be the first term and r be the common ratio of the GP. Sum of n terms of GP, Sn =... View Article
1 Plus 2 By 2 Fact Plus 3 Cube By 3 Fact Plus 4 Cube By 4 Fact Equals (1) 5e (2) 4e (3) 3e (4) 2e Solution: Given 1 + 2/2! + 33/3! + 43/4! + … Here tn = n3/n! = (n2-1)/(n-1)! + 1/(n-1)! =... View Article
The Value Of 2 By 1 Fact Plus 2 4 By 2 Fact 2 4 6 By 3 Is (1) e (2) 2e (3) 3e (4) none of these Solution: 2/1! + (2+4)/2! + (2+4+6)/3! +..... = 2/1! + 6/2! + 12/3! + 20/4! +........ (1)... View Article
If A Sigma N Eq 0 To Infinity B Eq Sigma 1 To Infinity X 3n 2 By 3n 2 Factorial And C Then The Value Of A3 B3 C3 3abc Is (1) 1 (2) 0 (3) -1 (4) 2 Solution: Given a = ∑n = 0∞ x3n/3n! b = ∑n = 1∞ x3n-2/(3n-2)! c = ∑n = 1∞ x3n-1/(3n-1)! Adding, we get... View Article