My best interpretation of this is that you're given a geometric progression consisting of [tex]n[/tex] terms, the first of which is [tex]a_1=7[/tex] and the last of which is [tex]a_n=1701[/tex]. (So we don't actually know right away how many terms there are.) The common ratio between terms is [tex]r=3[/tex]. You want to find the sum of all [tex]n[/tex] terms.
In a geometric progression, the [tex]k[/tex]-th term is determined by the previous term according to
[tex]a_k=ra_{k-1}[/tex]
Starting with [tex]a_1=7[/tex], we find
[tex]a_2=3a_1=21[/tex]
[tex]a_3=3a_2=3^2a_1=63[/tex]
[tex]a_4=3a_3=3^3a_1=189[/tex]
and so on. The general pattern for the [tex]k[/tex]-th term is then
[tex]a_k=3^{k-1}a_1=7\cdot3^{k-1}[/tex]
The last term in the sequence is [tex]a_n=1701[/tex], so
[tex]1701=7\cdot3^{n-1}\implies 243=3^5=3^{n-1}\implies n=6[/tex]
The sum of these [tex]n=6[/tex] terms is given by
[tex]S_6=\displaystyle\sum_{k=1}^6a_k=a_1+a_2+\cdots+a_6[/tex]
[tex]S_6=a_1+ra_1+r^2a_1+\cdots+r^5a_1[/tex]
Notice that
[tex]rS_6=ra_1+r^2a_1+r^3a_1+\cdots+r^6a_1[/tex]
[tex]\impleis S_6-rS_6=a_1-r^6a_1[/tex]
[tex]\implies(1-r)S_6=a_1(1-r^6)[/tex]
[tex]\implies S_6=a_1\dfrac{1-r^6}{1-r}[/tex]
With [tex]a_1=7[/tex] and [tex]r=3[/tex], we get a sum of
[tex]S_6=7\dfrac{1-3^6}{1-3}=2548[/tex]
