Question

AP of 1+3+5+...

Answer 1

it diverges

Answer 2

I think that is what you meant, the nth term

Answer 3

Tn= 1+2(n-1) = 2n-1

Answer 4

no, the nth term is (2n-1)

Answer 5

yeh I just figured that out lol

Answer 6

here is the actual question; S= log (a)+log (a^3/b)+log (a^5/b^2)+ log (a^7/b^3)+...

value of sum upto n terms is?

Answer 7

= log(a) + log(a^2) + log(a/b) + log(a^3) + log( (a^2/b^2)) + ....

Answer 8

that is the first three terms.

Answer 9

just by applying the law log(ab) = log(a)+log(b)

Answer 10

righto!

Answer 11

S= log(a)+ 2log(a) +log(a/b) + 3log(a) + 2log(a/b)

Answer 12

THATS by apply applying log(a^n) = n log(a)

Answer 13

please continue

Answer 14

then you get two arithmetic series.

Answer 15

You have

S= (log(a)+2log(a)+3log(a) + ......) + ( log(a/b) + 2log(a/b) + 3log(a/b) +....)

Answer 16

now, there will be "n" terms in the first summation, and "n-1" terms in the second summation.

Do you see this?

Answer 17

true

Answer 18

then you the sum of AP formula : S= (n/2) (2a +(n-1)d)

To sum both the series.

Answer 19

S= (n/2)(2log(a)+(n-1)log(a) ) + ((n-1)/2) ( 2log(a/b) + ( (n-1) -1) log(a/b) )

Answer 20

\[\S= \frac{n}{2}[2\log(a) +(n-1)\log(a) ] + \frac{n-1}{2}[2\log(\frac{a}{b}) + (n-2)\log(\frac{a}{b}) ]\]

Answer 21

then you can simplify it further.

Answer 22

why are there only 'n-1' terms in the second summation instead of 'n' terms?

Answer 23

The series starts at log(a). Every term after that is broken up into TWO terms.

The series for log(a/b) starts one term after the series of log(a).

Answer 24

hmm, i still didn't get it. i mean can't we say the series for log(a/b) commences with log(1/b).

Answer 25

Yes, it's kinda hard to explain over the internet but there is nothing much I can really do to explain it.

Answer 26

okay, so for a AP of odd terms, what would be 'n'?

Answer 27

notice the pattern , the log(a) series is "one term ahead" of the log(a/b) series

Answer 28

"AP of odds terms"?

Answer 29

Look at what I did above, it is correct (I am fairly sure).

Answer 30

RIGHT! it feels so liberating to finally get it! :)

Answer 31

yeah, AP of odd terms!

Answer 32

EDIT:

There was a typo. It should be :

The first term, we do nothing to it. log(a). The second term is broken into two terms: 2log(a) and log(a/b) The third term is broken into two terms,: 3log(a) and 2log(a/b)

Answer 33

There is no "AP of odd terms", what you are doing is summing two APs one with "n" terms and one with "n-1" terms.

Answer 34

The formula for the sum of an AP with "n" terms is \[\frac{n}{2}[2a+(n-1)d]\]

Answer 35

You use that to sum the first series!

Answer 36

so, generally speaking, sum of AP= n/2 (2a+(n-1)d) ; here, 'n' means the number of terms/last term?

Answer 37

Then for the second series, you have "n-1" terms so wherever you see an "n" in the sum formula, replace it with "n-1"

\[\S= \frac{n-1}{2}[2a+( (n-1) -1) d] \]

Answer 38

yeah, got it. please have a look at my above doubt

Answer 39

Yes the number

Answer 40

That is exacly what I did to get the answer like 10posts above.

Answer 41

haha! sorry for the trouble, but u were kind enough to clarify my trivial doubts