In a geometric progression, the second term has 9 less than the first term. The sum of the second and third terms is 30. Given that all the terms of the progression are positive, find the first term.

-Trying to solve this problem for awhile now. Also, am confused when to use Sn = a(r^n-1)/r-1 or Sn = a(1-r^n)/1-r. Any help would be appreciated, thanks.

Question

In a geometric progression, the second term has 9 less than the first term. The sum of the second and third terms is 30. Given that all the terms of the progression are positive, find the first term.

-Trying to solve this problem for awhile now. Also, am confused when to use Sn = a(r^n-1)/r-1 or Sn = a(1-r^n)/1-r. Any help would be appreciated, thanks.

anonymous · Answer

first term is a=x
second term, ar=x-9
third term, ar^2=30

r=x-9/x

a*r^2 shall give u the answer

yanasidlinskiy · Answer

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anonymous · Answer

@Saarthakjain
Hey, thanks for the reply. 
I don't understand how the third term is: ar^2=30? 
The question mentions the sum of the second and third terms = 30. 
Does that mean that S3=30-x? (x being the first term).

@YanaSidlinskiy 
Thanks!

yanasidlinskiy · Answer

Your Welcome:D

amistre64 · Answer

$$\frac{-1}{-1}*\frac{1-r^n}{1-r}=\frac{r^n-1}{r-1}$$
so your Sn formulas are the same thing.

amistre64 · Answer

the sum of the second and third terms is not properly a partial sum.  partial sums are when we add all the terms from 1 to N

amistre64 · Answer

a = a
ar = a-9
ar^2 = 39 - a

seems to be what we have to work with

anonymous · Answer

Yup, I got that far. Not sure how to go from there though.

amistre64 · Answer

justa few thoughts 

9 < a < 39

0 < a-9 < 30

and r < 1,  since ar < a

amistre64 · Answer

r = 1 - 9/a

ar = sqrt(a*ar^2)
    = sqrt(39a - a^2)

r = sqrt(39a - a^2)/a

since r=r

(a-9)/a = sqrt(39a - a^2)/a

a-9 = sqrt(39a - a^2)

(a-9)^2 = 39a - a^2

a^2 -18a +81 = 39a - a^2

2a^2 -57a +81 = 0

anonymous · Answer

Ok. I think I got it:
Assuming that the ratio is constant, 
$$(a-9)/(a) = (39-1)/(a-9)$$
$$a^2-18a+81 = 39a - a^2$$
$$2a^2-57a+81 = 0$$
$$\therefore a = 81$$

amistre64 · Answer

seems like we concur :)

anonymous · Answer

Awesome, thanks for the enlightenment!! :)

amistre64 · Answer

youre welcome, and good job as well :)

amistre64 · Answer

i get a=27

a = 27
ar = 18
ar^2 = 12

anonymous · Answer

Oops, sorry. I got 21 as well. Somehow I typed 81 instead.

amistre64 · Answer

i typo all the time, its these fat stubby fingers on this tiny little keyboard fer sure :)

anonymous · Answer

Haha, is there anyway to edit something I've posted?

amistre64 · Answer

no, theres not :)  you simply have to clarify it

amistre64 · Answer

you asked about trig in the message, any specific example you have to work with?