Geometric progression first term and common ratio of progression

Question

Do I simplify? Trying to find first term and common ratio of the progression aswell as the sum of all the terms in the progression for homework please help trying to know what I’m doing wrong so I can move on

Answer 1

$(3\cdot 2^{n-1})_{n=1}^{10}$ gives:

$3\cdot 2^{1-1}, 3\cdot 2^{2-1},\dotso, 3\cdot 2^{10-1}$

So let $a_n=3\cdot 2^{n-1}$. You are now asked to find $q$ such that $q\cdot a_n = a_{n+1}$

Answer 2

What does the notation $(3*2^{n-1})_{n=1}^{10}$ mean?

A geometric progression in general is usualy given as a sequence $a_{k}$ which satisfies $\frac{a_{k+1}}{a{k}}=constant$ which can be written as $a_{k+1}=a_{k}c\ \text{where c is the constant ratio}$.

If you start at given $a_{0}$ and with given constant c then the kth value calculates as $a_{0}c^{k-1}$.

So you want to calculate the first 10 elements(terms) for $a_{0}=3$ and $c=2$ ???

The first value of the geometric progression $3*2^n$ if n starts at 0 (or $3*2^{n-1}$ if n starts at 1) should be $3$,the second $2*3=6$,the 10th $3*2^9=3*64=192$ and so on..

The sum of your progression up to 10 calculates as $$3*\sum_{n=1}^{10}2^{n-1}=3*\frac{2^{10}-1}{2-1}=3*(2^{10}-1)$$