Let S1,S2,S3, ............ S10 respectively be the sum to 12 terms of 10 A.P. whose first terms are 1, 2, 3, ....... 10 and the common difference are 1, 3, 5, .........., 19 respectively. Then ∑i=110Si is equal to
(A) 7220 (B) 7380 (C) 7260 (D) 7360
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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