Loga[b] represents the logarithm based on b, and b is a real number.

"*" indicates the multiplication symbol.

""it means power.

First question

T < 10 >= 145 =(b & lt； 1 & gt； +b & lt； 10 >) 10/2, and T<n> is the sum of the first n items of B.

Then by B.

And b <10 > = b <1>; +9d(d is the tolerance of arithmetic progression b), so d=3.

In addition, the general formula of b is b < n >=3n-2.

Then a < n > = loga [1+1/b <; n & gt]=loga[(3n- 1)/(3n-2)]

That is, the general formula of a is a.

Compare S<n> with (1/3) loga [b

And 3s

= 3 loga[(2/ 1)(5/4)……((3n- 1)/(3n-2))]

= 3 loga[(2 * 5 *……*(3n- 1))/( 1 * 4 *……*(3n-2))]

=loga[(2*5*……*(3n- 1))/( 1*4*……*(3n-2))]^3

loga[b & lt； n+ 1 & gt； ]=loga[3n+ 1]

Look at the real numbers of the two first:

[(2*5*……*(3n- 1))/( 1*4*……*(3n-2))]^3

& gt[(2 * 5 *……*(3n- 1))/( 1 * 4 *……*(3n-2))]

*[(3 * 6 *……* 3n)/(2 * 5 *……*(3n- 1))]

*[(4 * 7 *……*(3n+ 1))/(3 * 6 *……* 3n)]

=[2 * 3 * 4 * 5 *……*(3n+ 1)/( 1 * 2 * 3 * 4 *……* 3n)]

=[3n+ 1]

(This step is the key just now, using zoom method.)

In other words, 3s

Look at the base, this is a:

There seems to be no range of the number A in the title, which needs to be discussed.

Note that the natural range of A is a>0 and A is not 1.

When 0

When a> is at 1, loga[x] is a monotonically increasing function, 3s.

So the conclusion is:

When 0

When a> is in 1, s

The second question {the first question}

General formula b of geometric series

e = b & lt 1 & gt； b & lt2 & gt……b & lt； 50 >；

= b & lt 1 & gt； * b & lt 1 & gt； q *……* b & lt； 1 & gt； q^49

=(b & lt； 1 & gt； ^50)*q^(25*49)

f = b & ltn-49 >b & ltn-48 >……b & lt； n & gt

= b & lt 1 & gt； q^(n-50)*b<； 1 & gt； q(n-49)*……* b & lt； 1 & gt； q^(n- 1)

=(b & lt； 1 & gt； ^50)*q^(25*(2n-5 1))

ef =(b & lt； 1 & gt； ^50)*q^(25*49)*(b<； 1 & gt； ^50)*q^(25*(2n-5 1))

=(b & lt； 1 & gt； ^ 100)*q^(50*(n- 1))

So the right side of the equation

(ef)^(n/ 100)=(b<； 1 & gt； ^n)*q^(n(n- 1)/2)

The left side of the equation again.

b & lt 1 & gt； b & lt2 & gt……b & lt； n & gt= b & lt 1 & gt； * b & lt 1 & gt； q *……* b & lt； 1 & gt； q^(n- 1)

=(b & lt； 1 & gt； N) * q (n (n- 1)/2) = right.

Therefore, the original equation holds.

The second question {the second question}

Just turn equal proportion into arithmetic, multiplication into addition and power into multiplication.

So the conclusion is:

In a positive arithmetic progression {b

b & ltn-49 >+b & lt； n-48 >+……+b & lt； N & gt=f, where n is a natural number greater than 49,

Then B.

{Prove} (Similar to the first question, just write a short step)

E = 50b

f = 50b & lt 1 & gt； +25*(2n-5 1)d

e+f = 100 b & lt； 1 & gt； +25*(2n-2)

(e+f)(n/ 100)= nb & lt； 1 & gt； +n(n- 1)/2 = b & lt； 1 & gt； +b & lt； 2 & gt+……+b & lt； n & gt

The conclusion is valid.