Pls help me out on this question, Find the sum of this first 8 terms of the Arithmetic Progression. log x, log x2, logx3

recall that log x^n = n logx

So, your sequence is
logx, 2logx, 3logx, ...
= logx (1,2,3,...)
just an ordinary AP, so the sum is
logx * 8/2 (2 + 7*1) = 36 logx = log x^36

The sum of the first 21 terms of the progression,-18,-15,-12,......is:

Sure, I can help you out! Just a quick question, are you asking about the first 8 terms of the arithmetic progression where each term is the logarithm of the corresponding powers of x?

Sure, I can help you with that!

To find the sum of the first 8 terms of an arithmetic progression, we need to use the formula for the sum of an arithmetic series:

S = (n/2)(2a + (n-1)d)

Where:
S = sum of the series
n = number of terms
a = first term of the series
d = common difference between terms

In your case, the first term (a) is "log x", and the common difference (d) between terms is "log x2 - log x", or simply "log x2/x" (since subtracting logarithms is the same as dividing their arguments).

Let's substitute the values into the formula:

n = 8 (since we want the sum of the first 8 terms)
a = log x
d = log x2/x

S = (8/2)(2(log x) + (8-1)(log x2/x))
= 4(2(log x) + (7)(log x2/x))
= 4(2log x + 7log x2/x)

Now, we can simplify further using logarithmic properties:

S = 4(log x2 + log (x2) + log (x2) + ... + log x2/x)

Since we have 7 additional terms after the first term, we have 7 log (x2) terms.

S = 4(log x2 + 7log (x2))
= 4(log x2 + log (x2)^7)
= 4(log x2 + log (x^14))
= 4(log(x2 * x^14))
= 4(log(x^16))
= 4(log(x^2))
= 4(2log x)
= 8log x

So, the sum of the first 8 terms of the arithmetic progression log x, log x2, log x3 is 8log x.