The Sum Of 100 Terms Of The Series 09 Plus 009 0009 Will Be Equal To
(1) 1- (1/10)100 (2) 1+ (1/10)106 (3) 1- (1/10)106 (4) 1+ (1/10)100 Solution: Here first term, a = 0.9 Common... View Article
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If The Nth Term Of A Gp 5 5 By 2 5 By 4 Is 5 By 1024 Then The Value Of N Is
1) 11 2) 10 3) 9 4) 4 Solution: Let a be first term of GP with common ratio r.... View Article
The 6th Term Of A Gp Is 32 And Its 8th Term Is 128 Then The Common Ratio Of The Gp Is
1) -1 2) 2 3) 4 4) -4 Solution: Let a be first term of GP with common ratio r.... View Article
If The 10th Term Of A Geometric Progression Is 9 And 4th Term Is 4 Then Its 7th Term Is
(1) 6 (2) 36 (3) 4/9 (4) 9/4 Solution: Let a be the first term of GP with common ratio... View Article
If The Pth Qth And Rth Term Of A Gp Are A B C Respectively Then Aq Rbr P Cp Q Is Equal To
(1) 0 (2) 1 (3) abc (4) pqr Solution: Let A be first term of GP with common ratio R.... View Article
In A Geometric Progression Consisting Of Positive Terms Each Term Equals The Sum Of The Next Two Terms
(1) (1-√5)/2 (2) √5/2 (3) √5 (4) (√5-1)/2 Solution: Let a, ar, ar2 be three terms of GP. Given a... View Article
If P Q Th Term Of A Gp Be M And P Qth Term Be N Then The Pth Term Will Be
(1) m/n (2) mn (3) √(mn) (4) 0 Solution: Let a be first term of GP with common ratio r.... View Article
If The 4th 7th And 10th Terms Of A Gp Be A B C Respectively Then The Relation Between A B C Is
(1) b = (a+c)/2 (2) a2 = bc (3) b2= ac (4) c2 = ab Solution: Let A be first... View Article
Sum Of N Terms Of The Following Series 1 Cube 3 Cube 5 Cube Is
(1) n2(2n2 – 1) (2) n3(n – 1) (3) n3 + 8n + 4 (4) 2n4 + 3n2 Solution: Given... View Article
7th Term Of The Sequence Root 2 Root 10 5 Root 2 Is
(1) 125√10 (2) 25√2 (3) 125 (4) 125√2 Solution: Here a = √2 Common ratio, r = √5 nth term... View Article
The Nth Term Of The Series 2 By 1 Fact 7 By 2 Fact Is
(1) n(3n-1)/2(n)! (2) n(3n+1)/2(n)! (3) n(3n)/2(n)! (4) none of these Solution: Given series 2/1! + 7/2! + 15/3! + 26/4!... View Article
11 Sq Plus 12 Sq Plus 13 Sq Plus 20 Sq Equal
1) 2481 2) 2483 3) 2485 4) 2487 Solution: Sum of squares n natural numbers = n(n+1)(2n+1)/6 112 + 122... View Article
First Term Of 11th Term Of The Following Groups 1 2 3 4 5 Is
(1) 89 (2) 97 (3) 101 (4) 123 Solution: Given groups (1), (2,3,4), (5,6,7,8,9)… In each group the numbers are... View Article
The Sum Of N Terms Of 2n 1 Plus 2 2n 3 Plus 3 2n 5 Is Eq
(1) (n+1)(n+2)/6 (2) n(n+1)(n+2)/6 (3) n(n+1)(2n+3) (4) n(n+1)(2n+1)/6 Solution: Given series (2n-1) + 2(2n-3) + 3(2n-5) +…n terms Here tr... View Article
If Nth Term Of A Series Be 3 N N 1 Then The Sum Of N Terms Of The Series Is
(1) (n2+n)/3 (2) (n3+8n)/3 (3) (n2+8n)/5 (4) (n2-8n)/3 Solution: Given nth term of the series, tn = 3 + n(n-1)... View Article
11 Cube Plus 12 Cube 20 Cube
(1) Is divisible by 5 (2) Is an odd integer divisible by 5 (3) Is an even integer which is... View Article
The Sum Of N Terms Of The Following Series 1 2 Plus 2 3 Plus 3 4 Plus 4 5 Shall Be
(1) n3 (2) n(n+1)(n+2)/3 (3) n(n+1)(n+2)/6 (4) n(n+1)(2n+1)/3 Solution: 1.2 + 2.3 + 3.4 + 4.5 +…n terms Here tn... View Article
1 Cube Plus 2 Cube Plus 12 Cube By 1 Sq 2 Sq Plus 12 Sq Equal
(1) 234/25 (2) 243/35 (3) 263/27 (4) none of these Solution: Sum of cubes of n natural numbers = [n(n+1)/2... View Article
The Sum Of The Series 1 3 Square 2 Plus 2 5 Sq Upto 20 Terms Is
(1) 188090 (2) 189080 (3) 199080 (4) none of these Solution: Let S= 1.32 + 2.52 +3.72 + …upto 20... View Article
Sum Of The Squares Of First N Natural Numbers Exceeds Their Sum By 330
(1) 8 (2) 10 (3) 15 (4) 20 Solution: Sum of n natural numbers = n(n+1)/2 Sum of squares n... View Article
