Infinite GP
Trending Questions
Q. For k∈N, let 1α(α+1)(α+2)…(α+20)=20∑k=0Akα+k, where α>0. Then the value of 100(A14+A15A13)2 is equal to
Q. Let Δr=∣∣
∣
∣∣r−1n6(r−1)22n24n−2(r−1)33n33n2−3n∣∣
∣
∣∣. Then the value of n∑r=1Δr is:
- 1
- 3
- 2
- 0
Q.
The term of the series .is
Q. If 114+124+134+⋯upto ∞=π490, then the value of 114+134+154+⋯upto ∞ is
- π445
- π496
- π4124
- π4196
Q. In an infinite G.P., if the first term is x and sum of all terms is 5, then
- x∈(0, 10)
- x∈(−∞, 0)
- x∈(10, ∞)
- x∈[1, 10]
Q. If x, y, z are in G.P. and x+y, y+z, z+x are in A.P., where x≠y≠z, then common ratio of the G.P. is
- 4
- −1
- 2
- −2
Q. Find the sum of the following geometric series:
(x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+⋯ to n terms;
(x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+⋯ to n terms;
Q. If 11+103+1005+⋯n terms=10n+1+xn2+y9, then the value of x−y is
- 19
- 1
- −1
- −19
Q. The sum of infinite terms of a G.P. is x and on squaring the each term of it, the sum will be y, then the common ratio of this series is
- x2−y2x2+y2
- x2+y2x2−y2
- x2−yx2+y
- x2+yx2−y
Q. The value of 1∫0xa−1lnxdx, (where a is parameter) is equal to
- ln|a−1|
- ln|a+1|
- aln|a+1|
- aln|a−1|
Q. Let a+ar1+ar12+... and a+ar2+ar22+... be two infinite series of positive numbers with the same first term, where r1, r2<1. The sum of the first series is r1 and the sum of the second series is r2 then the value of r1+r2 is
Q. The sum of the series ∞∑n=1n2+6n+10(2n+1)! is equal to :
- 418e+198e−1−10
- −418e+198e−1−10
- 418e−198e−1−10
- 418e+198e−1+10
Q. If ∞∑k=11(k+2)√k+k√k+2=√a+√b√c, where a, b, c∈N and a, b, c∈[1, 15], then a+b+c is equal to
Q. If x, y, z are in G.P. and x+y, y+z, z+x are in A.P., where x≠y≠z, then common ratio of the G.P. is
- 4
- −1
- 2
- −2
Q. ∞∑i=1∞∑j=1∞∑k=11ai+j+k is equal to (where |a|>1)
- (a−1)−3
- 3a−1
- 3a3−1
- None of these
Q.
The product of infinite terms in x12.x14.x18...........∞ is
0
infinity
1
x
Q. If f:[1, ∞)→B defined by the function f(x)=x2−2x+6 is a surjection, then B is equals to
- [1, ∞)
- [6, ∞)
- [5, ∞)
- [2, ∞)
Q.
If sum of infinite terms of a G.P. is 3 and sum of squares of its terms is 3, then its first term and common ratio are
[RPET 1999]
3/2, 1/2
1, 1/2
3/2, 2
None of these
Q.
Match the statements of Column I with values of Column II
Column IColumn II(A) ∫e2x−2exe2x+1dx=A ln(e2x+1)+B tan−1(ex)+c(p) A=−12, B=−14(B) ∫√x+√x2+2dx=A{x+√x2+2}32+B√x+√x2+2+c(q) A=12, B=−2(C) ∫cos 8x−cos 7x1+2 cos 5xdx=A sin 3x+B sin 2x+c(r) A=13, B=−2(D) ∫ln xx3dx=Aln xx2+Bx2+c(s) A=13, B=−12
A-s , B-r , C-q , D-p
A-p , B-s , C-r , D-q
A-p , B-q , C-r , D-s
A-q , B-r , C-s , D-p
Q. (625)6.25×(25)2.6(625)6.75×(5)1.2=?
- 5
- 10
- 15
- 25
Q. Let Δr=∣∣
∣
∣∣r−1n12(r−1)22n28n−4(r−1)33n36n2−6n∣∣
∣
∣∣. Then the value of n∑r=1Δr is:
- 1
- 0
- 2
- 3
Q. If 1∫0ex2(x−α)dx=0, then which of the following is true for α?
- 1<α<2
- α<0
- 0<α<1
- α=0
Q. Minimise and maximise z=5x+2y subject to the following constraints :
x−2y≤2
3x+2y≤12
−3x+2y≤3
x≥0, y≥0
x−2y≤2
3x+2y≤12
−3x+2y≤3
x≥0, y≥0
Q. If (1−y)(1+2x+4x2+8x3+16x4+32x5)=1−y6, where y≠1, x≠12, then the value of yx is
- 12
- 2
- 2524
- 2425
Q. ∑ni=1∑ij=1∑jk=11 is equal to
- n(n+1)2
- [n(n+1)2]2
- n(n+1)(n+2)6
- n(n+1)(2n+1)6
Q. 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
[AIEEE 2002]
[AIEEE 2002]
5
3/5
- 8/5
- 1/5
Q. Evaluate: ∫x2x4+x2−2dx.
Q. Evaluate the definite integral ∫206x+3x2+4dx
Q. If (1−y)(1+2x+4x2+8x3+16x4+32x5)=1−y6, where y≠1, x≠12, then the value of yx is
- 12
- 2
- 2524
- 2425
Q. The value of x in (0, π) which satisfy the equation
8−+|cosx|+cos2x+|cos3x|+⋯to∞=43 is
8−+|cosx|+cos2x+|cos3x|+⋯to∞=43 is
- {π2, 3π4}
- {π4, 3π4}
- {π3, 2π3}
- {π6, 5π6}