Let the first term a and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to
(A) 231 (B) 241 (C) 210 (D) 220
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Let a1, a2, a3, .., an be n positive consecutive terms of an arithmetic progression. If d>0 is its common difference, then
limn→∞dn(1a1+a2+1a2+a3+...+1an-1+an) is
(c) : We have, a1, a2, ...., an are in A.P.
∴ a2- a1= a3- a2=......= an- an-1=d
Now, limn→∞dn(1a1+a2+1a2+a3+...+1an-1+an)
= limn→∞dn(a2-a1a2-a1+a3-a2a3-a2+...+an-an-1an-an-1)
= limn→∞dn(a2-a1d+a3-a2d+...+an-an-1d)
=limn→∞1nd(an-a1)
=limn→∞1nd(an-a1an+a1)=limn→∞1nd(n-1)dan+a1
=limn→∞1ndn(1-1n)dn(a1n+(1-1n)d+a1n)
=limn→∞(1-1n)dd(a1n+d-dn)+a1n=dd·d=1
Among
(S1): limn→∞1n2(2+4+6+...+2n)=1
(S2): limn→∞1n16(115+215+315+....+n15)=116
(d) : (S1) : limn→∞1n2(2+4+6+...+2n)
=limn→∞1n2×2(1+2+3+...+n)
=limn→∞1n2×2×n(n+1)2=limn→∞(1+1n)=1
∴ (S1) is true.
(S2) : limn→∞1n16(115+215+315+....+n15)=limn→∞1n16∑r=1nr15
=limn→∞1n∑r=1n(rn)15=∫01x15dx=|x1616|01=116
(S2) is also true.
limn→∞{(212-213)(212-215)....(212-212n+1)} is equal to
(c) : Let L = limn→∞{(212-213)(212-215)....(212-212n+1)}
By Sandwich Theorem,
(212-213)n<(212-213)(212-215)(212-217)....(212-212n+1)<(212-212n+1)n
⇒limn→∞ (212-213)n<L<limn→∞ (212-212n+1)n
As, limn→∞ (212-213)n=0 and limn→∞ (212-212n+1)n=0
∴ L = 0
limx→0((1-cos2(3x)cos3(4x))(sin3(4x)(loge(2x+1))5)) is equal to _____ .
(d) : limx→0[((1-cos23x)cos3(4x))(sin3(4x)(loge(2x+1))5)]
=limx→0[sin2(3x)cos34x×sin3(4x)[loge(2x+1)]5]
=limx→0[sin2(3x)(3x)2×sin3(4x)(4x)3×(3x)2×(4x)3cos3(4x)·[loge(2x+1)2x]5×(2x)5]
=9×6432=18
If α>β>0 are the roots of the equation ax2+bx+1=0, and limx→1α(1-cos(x2+bx+a)2(1-αx)2)12=1k(1β-1α), then k is equal to
(d) : Given, ax2+bx+1 = 0 has roots α, β, then x2+bx+a=0 has roots 1α,1β.
Now, limx→1α (1-cos(x2+bx+a)2(1-αx)2)1/2
=limx→1α (2sin2(x2+bx+a2)2(1-αx)2)1/2
=limx→1α (2sin2((x-1α)(x-1β)2)2(1-αx)2)1/2
=limx→1α (sin2((1-αx)(1-βx)2αβ)((1-αx)(1-βx)2αβ)2×((1-αx)(1-βx)2αβ)2(1-αx)2)1/2
=limx→1α (1-βx2αβ)=(α-β2α2β)=12α(1β-1α)
=1k(1β-1α) (⇒Given)
So, k = 2α