Irvin Dukes

2021-12-29

What is the 4th partial sum for the geometric sequence where
a) ${S}_{4}=-18$
b) ${S}_{4}=288$
c) ${S}_{4}=152$
d) ${S}_{4}=270$

Becky Harrison

Given data is
${a}_{1}=18$
common ratio $\left(r\right)=2$
We have to calculate 4-th partial sum of CnP.
$S={a}_{1}\left(\frac{1-rn}{1-r}\right)$
By putting the values in formulae.
$⇒S=18\left(\frac{1-{2}^{4}}{1-2}\right)$
$⇒S=18\left(\frac{1-16}{1-2}\right)$
$⇒S=18\left(\frac{-15}{-1}\right)$
$⇒S=18×15=270$

deginasiba

We have, sum to n terms of a geometric sequence is given by.

So, putting given values $n=4$, we get
${S}_{4}=\frac{18\left[{\left(2\right)}^{4}-1\right]}{2-1}=\frac{18\left(16-1\right)}{1}=18×15=270$
Hence, 4th partial sum $={S}_{4}=270$
Option c is correct

user_27qwe

The 4th partial sum of the sequence is the sum of the first 4 terms:
${S}_{4}=a\left(1\right)+a\left(2\right)+a\left(3\right)+a\left(4\right)$
Because the sequence is geometric, any term can be written in terms of the previous one:
a(n) = r a(n - 1) and, more importanly, in terms of the first term a(1), so that
$a\left(2\right)=ra\left(1\right)$
$a\left(3\right)=ra\left(2\right)={r}^{2}a\left(1\right)$
$a\left(4\right)=ra\left(3\right)={r}^{3}a\left(1\right)$
Then the 4th partial sum is
${S}_{4}=a\left(1\right)+ra\left(1\right)+{r}^{2}a\left(1\right)+{r}^{3}a\left(1\right)$
${S}_{4}=a\left(1\right)\left(1+r+{r}^{2}+{r}^{3}\right)$
With a(1) = 18 and r = 2 we have
${S}_{4}=18\left(1+2+4+8\right)=270$

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