Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if A5 = 5A3, then s9s5=______ ．

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if A5 = 5A3, then s9s5=______ ．

∵ {an} is an arithmetic sequence, S9 = a1 + A2 + +a9=9a5，S5=a1+a2+… +A5 = 5A3, s9s5 = 9a55a3 = 9, so the answer is 9

Let the sum of the first n terms of the arithmetic sequence [an] be SN. If A5 = 5A3, what is S9 / S5 equal to?

S9/S5
=2S9/2S5
=(a1+a9)*9/[(a1+a5)*5]
=9a5/5a3
=45a3/5a3
=9

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if A5 = 5A3, then s9s5=______ ．

∵ {an} is an arithmetic sequence, S9 = a1 + A2 + +a9=9a5，S5=a1+a2+… +A5 = 5A3, s9s5 = 9a55a3 = 9, so the answer is 9

Let the sum of the first n terms of the arithmetic sequence {an} be SN. If A5 = 5A3, then S9 / S5=____ Is this question wrong? How can A5 = 5A3? I hope those who can solve it can reply as soon as possible!
This problem comes from the "high school mathematics core problem solving methods and skills" edited by Zhang Yonghui!

No mistake, S9 / S5 = 9
A5=5A3
S9/S5=9*(A1+A9)/5*(A1+A5)=9*2(A5)/5*2(A3)=9*(A5)/_ 5*(A3)=9

In the equal ratio sequence {an}, q = 2, log2a1 + log2a2 +... + log2a10 = 25, then what is a1 + A2 +... + A10?
Please give the process

Log2a1 + log2a2 +... + log2a10 = log2 (A1 * A2 *... * A10) = 25 because A1 * A2 *... * A10 = A1 * a1q * a1q ^ 2... A1q ^ 9 = (A1) ^ 10 * q ^ 45, so log2a1 + log2a2 +... + log2a10 = log2 [(A1) ^ 10 * q ^ 45] = log2 (A1) ^ 10 + log2 Q ^ 45 = 10log2 a1 + 45log2 Q because q = 2

In the equal ratio sequence {an}, log2a1 + log2a2 +. + log2a10 = 35
Q = 2 for a1 + A2 +..... +a10

log2a1+log2a2+.+log2a10=35
=>a1a2…… a10=2^35
a1^10*2^((1+10)*10/2)=2^35
a1=1/4
a1+a2+…… +a10=a1(1-q^10)/(1-q)=(1/4)(1-2^10)/(1-2)=1023/4

In {an}, q = 2, log2a1 + log2a2 + +Log2a10 = 25, then a1 + A2 + +A10 equals ()
A. 237B. 10214C. 10234D. 250

According to the operation property of logarithm, log2a1 + log2a2 + +log2a10=log2（a1a2a3… a9a10）=log2（a1a10）5=25，∴（a1a10）5=225，∵q=2，∴a1=14，∴a1+a2+… +A10 = 14 (1 − 210) 1 − 2 = 10234, so C

Q = 2 log2a1 + log2a2 + log2a3 +... + log2a10 = 25, then a1 + A2 + a3 +. + a10=

log2a1+log2a2+log2a3+...+log2a10
=log2(a1*a2*...*a10)
=log2(a1^10*q^45)
=10*log2a1+45*log2q=25
log2a1=(25-45)/10
a1=1/4
a1+a2+a3+.+a10
=a1*(q^0+q^1+...+q^9)
=a1*(1-q^10)/(1-q)
=1/4*(2^10-1)
=2^8-1/4
=256-0.25
=255.75

In {an}, if an ＞ 0 and a3a4 = 4, then log2a1 + log2a2 + +Log2a6 value is ()
A. 5B. 6C. 7D. 8

In the equal ratio sequence, a3a4 = a1a6 = a2a5 = 4, log2a1 + log2a2 + +log2a6=log2（a1a2… A 6) = log 226 = 6

If an > 0, A3 * A4 = 4, then log2a1 + log2a2 +. Log2a6 =?

a3a4=(a1q)²(a1q³)=a1²q^5=4
log2(a1)+log2(a2)+...+log2(a6)
=log2(a1·a2·...·a6)
=log2(a1^6q^15)
=log2[(a1²q^5)³]
=3log2(4)
=3×2
=6