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Answer:
(b) s > t, a101 < b101
Step-by-step explanation:
log e (b1), log e (b2), log e (b3)...log e (b101) are in arithmetic progression
Know that, in arithmetic progression, the difference will be the same. So if we subtract the consecutive terms, they will equal one another. I take this is an example:
log e (b2) - log e (b1) = log e (b3) - log e (b2) = 2 [common difference log e 2]
=> log e (b2/b1) = log e (b3/b2)
=> b2/b1 = b3/b2 = b4/b3 = b5/b4... = 2 [common difference is log e 2]
Take b3/b2 as an example. b3/b2 = 2, b3 = 2*b2 = 2*2*b1 = 2^2(b1).
Take b4/b3. b4/b3 = 2, b4 = 2*b3 = 2*2*b2 = 2*2*2*b1 = 2^3b1.
Notice a pattern? bn = 2^(n-1) * b1
use geometric series with common ratio r^nth term [2^51] and first term a [a1 = 1]:
t = b1 + b2...b51 = b1 + 2b1...2^50b1
= b1 * (2^51 - 1) / 1
= b1(2^51 - 1)
use sum of terms in arithmetic sequence where n = number of terms:
s = a1 + a2...a51 = (a1 + a51)/2 * n = (b1 + b51)/2 * n
= (b1 + b51)/2 * 51
= (b1 + 2^50)/2 * 51 = b1(51/2(1 + 2^50))
Now since (2^51-1) is clearly less than 51/2 * (1 + 2^50), s > t
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This is how to determine the nth term of an arithmetic sequence:
a101 = a1 + (101 - 1)d
= b1 + 100 * (b51 - b1)/50 = b1 + 2b51 - 2b1 = 2b51 - b1 = b1(2^51 - 1)
b101 = 2^100 * b1
Since 2^100 > 2^51 - 1, b101 > a101