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5 Years Ago

If The Sum Of The Series 1+2/x + 4/x^2 + 8/x^3 +.....infinite Number Then

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5 Years Ago

If The Sum Of The Series 1+2/x + 4/x^2 + 8/x^3 +.....infinite Number ThenÂ

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5 Years Ago

If S Is The Sum To Infinity Od G.P., Â whose Terms Is A , Then The Sum Of The First N Terms Is

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5 Years Ago

Let Sn Denotes The Sum Of N Terms Of An A.P. If S2n = 3Sn The Ratio S3n / Sn=

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If A,b,c Are Three Unequal Number Such That A,b,c Are In A.P. And B-a , C-b, A Are In G.P. , Then A:b:c Is

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Modified 5 Year Ago
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5 Years Ago

Â If A,b,c Are In A.P. And A^2, B^2, C^2 Are In H.P. , Then

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5 Years Ago

If A And G Are Arithmetic And Geometric Mean And X^2-2Ax=G^2=0, Then

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5 Years Ago

Jairam Purchased A House In Rs. 15000 And Paid Rs. 5000 At Once. Rest Money He Promised To Pay In Annual Installment Of Rs.1000 With 10% Per Annum Interest. How Much Money Is To Be Paid By Jairam

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Modified 5 Year Ago
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102 views
5 Years Ago

The Harmonic Mean Of Two Numbers Is 4 And The Arithmetic And Geometric Means Satisfy The Relation 2A+G^2=27 , The Numbers Are

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5 Years Ago

If The Roots Of The Equation X^3 -12x^2+39x-28=0 Are In A.P., Then Their Common Difference Will Be

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5 Years Ago

If 100 Times The 10thÂ tern Of An A.P. With Non Zero Common DifferenceÂ Equal The 50 Times Its 50thÂ terms Then The 150thÂ term Of This AP Is

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5 Years Ago

If P And Q Are Positive Real Numbers Such That The Maximum Value Of (p+q) Is

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5 Years Ago

2^1/4 .4^1/8 .8^1/16 .16^1/32...is Equal To

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Modified 5 Year Ago
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5 Years Ago

Sum Of Infinite Number Of Terms In G.P. Is 20 And Sum Of Their Square Is 100. The Common Ratio Of G.P. Is

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Modified 5 Year Ago
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5 Years Ago

The Sum Of All Possible Product Of The First N Natural Number Taken Two By Two Is

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Modified 5 Year Ago
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5 Years Ago

The N^th Terms Of Series 1/1 + (1+2)/2 + (1+2+3)/3 +...........will Be

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5 Years Ago

2+4+7+11+16+...to N Terms=

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Modified 5 Year Ago
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5 Years Ago

Sum Of First N Terms In The Following Series Cot^-1 3+cot^-1 7 + Cot ^-1 13 + Cot^-1 21 + ...is Given By

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Modified 5 Year Ago
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5 Years Ago

The Sum Of The Series 1^2.2 + 2^2.3 + 3^2.4 + ......to N Terms Is

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Modified 5 Year Ago
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5 Years Ago

The Sum Of N Terms Of The Series Whose N^th Term Is N(n+1) Is Equal To

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Modified 5 Year Ago
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5 Years Ago

Sum Of The N Terms Of The Series 3/1^2 + 5/ (1^2 +2^2) + 7/(1^2+2^2+3^2) +.....is

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Modified 5 Year Ago
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5 Years Ago

1^3+2^3+3^3+4^3+5^3......+12^3 / 1^2 + 2^2+3^2+4^2+......+12^2 =

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Modified 5 Year Ago
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5 Years Ago

The Sum Of (n-1) Terms Of 1+(1+3) + (1+3+5)+.....is .

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5 Years Ago

The Sum Of N Terms Of The Series 1+3x+6x^2+10x^3 +........infinite Will Be

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Modified 5 Year Ago
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5 Years Ago

The Sum Of N Terms Of The Series 3/1^.2^2 + 5/2^2.3^2 + 7/ 3^2.4^ +.....is

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Modified 5 Year Ago
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5 Years Ago

If |x| Less 1 , Then The Sum Of The Series 1+2x+3x^2+4x^3+..â€¦......infinite Will Be

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Modified 5 Year Ago
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5 Years Ago

The Sum Of The Series 1.2.3.4+3.4.5+.....to N Terms Is

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Modified 5 Year Ago
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5 Years Ago

N^th Term Of The Series 2+4+7+11+.â€¦...will Be

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Modified 5 Year Ago
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5 Years Ago

If A,b,c,d Are In G.P , Then (a+b+c+d)^2 Is Equal To

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Modified 5 Year Ago
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5 Years Ago

In A G.P. , T2+t5=216 And T4:t6 = 1:4 And All Terms Are Integers, Then It First Term Is

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Modified 5 Year Ago

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3416 of 46300 Questions

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If The Sum Of The Series 1+2/x + 4/x^2 + 8/x^3 +.....infinite Number Then

If The Sum Of The Series 1+2/x + 4/x^2 + 8/x^3 +.....infinite Number ThenÂ

If S Is The Sum To Infinity Od G.P., Â whose Terms Is A , Then The Sum Of The First N Terms Is

Let Sn Denotes The Sum Of N Terms Of An A.P. If S2n = 3Sn The Ratio S3n / Sn=

If A,b,c Are Three Unequal Number Such That A,b,c Are In A.P. And B-a , C-b, A Are In G.P. , Then A:b:c Is

Â If A,b,c Are In A.P. And A^2, B^2, C^2 Are In H.P. , Then

If A And G Are Arithmetic And Geometric Mean And X^2-2Ax=G^2=0, Then

Jairam Purchased A House In Rs. 15000 And Paid Rs. 5000 At Once. Rest Money He Promised To Pay In Annual Installment Of Rs.1000 With 10% Per Annum Interest. How Much Money Is To Be Paid By Jairam

The Harmonic Mean Of Two Numbers Is 4 And The Arithmetic And Geometric Means Satisfy The Relation 2A+G^2=27 , The Numbers Are

If The Roots Of The Equation X^3 -12x^2+39x-28=0 Are In A.P., Then Their Common Difference Will Be

If 100 Times The 10thÂ tern Of An A.P. With Non Zero Common DifferenceÂ Equal The 50 Times Its 50thÂ terms Then The 150thÂ term Of This AP Is

If P And Q Are Positive Real Numbers Such That The Maximum Value Of (p+q) Is

2^1/4 .4^1/8 .8^1/16 .16^1/32...is Equal To

Sum Of Infinite Number Of Terms In G.P. Is 20 And Sum Of Their Square Is 100. The Common Ratio Of G.P. Is

The Sum Of All Possible Product Of The First N Natural Number Taken Two By Two Is

The N^th Terms Of Series 1/1 + (1+2)/2 + (1+2+3)/3 +...........will Be

2+4+7+11+16+...to N Terms=

Sum Of First N Terms In The Following Series Cot^-1 3+cot^-1 7 + Cot ^-1 13 + Cot^-1 21 + ...is Given By

The Sum Of The Series 1^2.2 + 2^2.3 + 3^2.4 + ......to N Terms Is

The Sum Of N Terms Of The Series Whose N^th Term Is N(n+1) Is Equal To

Sum Of The N Terms Of The Series 3/1^2 + 5/ (1^2 +2^2) + 7/(1^2+2^2+3^2) +.....is

1^3+2^3+3^3+4^3+5^3......+12^3 / 1^2 + 2^2+3^2+4^2+......+12^2 =

The Sum Of (n-1) Terms Of 1+(1+3) + (1+3+5)+.....is .

The Sum Of N Terms Of The Series 1+3x+6x^2+10x^3 +........infinite Will Be

The Sum Of N Terms Of The Series 3/1^.2^2 + 5/2^2.3^2 + 7/ 3^2.4^ +.....is

If |x| Less 1 , Then The Sum Of The Series 1+2x+3x^2+4x^3+..â€¦......infinite Will Be

The Sum Of The Series 1.2.3.4+3.4.5+.....to N Terms Is

N^th Term Of The Series 2+4+7+11+.â€¦...will Be

If A,b,c,d Are In G.P , Then (a+b+c+d)^2 Is Equal To

In A G.P. , T2+t5=216 And T4:t6 = 1:4 And All Terms Are Integers, Then It First Term Is

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