Sequences and Series
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New answer posted
2 months agoContributor-Level 9
S? =0 ⇒ (a? +a? ) (11/2)=0 ⇒ a? =-a?
2a? +10d=0 ⇒ a? =-5d.
Sum = a? +a? +.+a? = (a? +a? ) (12/2) = 6 (2a? +22d)
= 6 (2a? +22 (-a? /5) = 6 (2a? -22a? /5) = 6 (-12a? /5)=-72a? /5. k=-72/5.
New answer posted
2 months agoContributor-Level 10
Let three terms of G.P. are a/r, a, ar product=27
⇒ a³=27 ⇒ a=3
S=3/r+3r+3
For r>0
(3/r+3r)/2 ≥ √9 = 3
⇒ 3/r+3r≥6
For r<0, 3/r+3r-6
From (1) and (2)
S∈ (-∞, -3]∪ [9, ∞)
New answer posted
2 months agoContributor-Level 10
α and β are roots of 5x²+6x-2=0
⇒ 5α²+6α-2=0
⇒ 5α? ²+6α? ¹-2α? =0
(By multiplying α? )
Similarly 5β? ²+6β? ¹-2β? =0
By adding (1) and (2)
5S? +6S? -2S? =0
For n=4
5S? +6S? =2S?
New answer posted
2 months agoContributor-Level 9
α, β are roots of x²+5√2x+10=0. α+β=-5√2, αβ=10.
P? + 5√2P? + 10P? = 0.
So P? +5√2P? =-10P? P? +5√2P? =-10P?
Expression = P? (-10P? )/P? (-10P? ) = 1.
New answer posted
2 months agoContributor-Level 9
S? = 3n/2 [2a+ (3n-1)d]. S? = 2n/2 [2a+ (2n-1)d].
S? =3S? ⇒ 3n/2 [2a+ (3n-1)d] = 3 (2n/2) [2a+ (2n-1)d].
2a+ (3n-1)d = 2 [2a+ (2n-1)d] ⇒ 2a+ (n-1)d=0.
S? /S? = (4n/2 [2a+ (4n-1)d]) / (2n/2 [2a+ (2n-1)d]) = 2 [- (n-1)d+ (4n-1)d]/ [- (n-1)d+ (2n-1)d] = 2 (3n)/ (n)=6.
New answer posted
2 months agoContributor-Level 9
Sol. Σ? k(k+1)/2 = (1/2)Σ(k²+k) = (1/2)[ (5051101/6) + (5051/2) ]
The solution uses k=1 to 20.
(1/2)[ (202141/6) + (2021/2) ] = (1/2)[2870+210] = 1540
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