Question

find the nth term and sum of n terms of the series 10+12+36+108++324....​

Answer 1

Answer:

Like the other answers have already explained, each term in the series is the previous term, times three. This series is called a geometric series, and each term can be described as the following:

an=a1rn−1  

an  being the nth term,  a1  being the first term, and  r  being the common ratio (the current term divided by the previous term).

In this case, each term can be described by

an=4∗3n−1  

where

a1=4∗30=4∗1=4  

a2=4∗31=4∗3=12  

And so on.

Now, we need to find a convenient way to add  n  terms.

Let  Sn  be the sum of  n  terms in a geometric series.

Sn=a1+a2+...+an−1+an  

Then, if we substitute each term for  an=a1rn−1  we find that

Sn=a1+a1r+a1r2+...+a1rn−2+a1rn−1  

Next, try taking the entire series, and multiply it by  r .

rSn=a1r+a1r2+a1r3+...+a1rn−1+a1rn  

Then, try subtracting  rSn  from  Sn . All the terms cancel, except for  a1  from  Sn  and  a1rn  from  rSn.  

Sn−rSn=a1−a1rn  

Factor out  Sn  on the left side and  a1  on the right side.

Sn(1−r)=a1(1−rn)  

Isolate  Sn .

Sn=a1(1−rn)(1−r)  

Now that that’s over with, let’s address the problem at hand. We know that  Sn  is equal to  4372 , we know that our first term is  4 , and we know our common ratio is  3 . Plugging that into our formula that we just derived, we find the that

4372=4(1−3n)(1−3)  

4372=4(1−3n)−2  

−2(4372)=4(1−3n)  

−2(4372)4=1−3n  

−2(4372)4−1=−3n  

−2187=−3n  

(Remember,  −3n=−(3n) , not  (−3)n )

2187=3n  

37=3n  

n=7  

Tada!

Step-by-step explanation: