What is the sum of the first seven terms of the geometric series 3 − 6 + 12 − . . . ?

Hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r ≠ 1, where a1 is the first term and r is the common ratio.

S7 = −63
S7 = −9
S7 = 33
S7 = 129

Question

What is the sum of the first seven terms of the geometric series 3 − 6 + 12 − . . . ?

Hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r ≠ 1, where a1 is the first term and r is the common ratio.

S7 = −63
 S7 = −9
 S7 = 33
 S7 = 129

OliveSpice89 · Answer

Ok so what do you believe it is?

snowflake0531 · Answer

This is adding/subtracting the double of the previous number
3-6+12-24+48-96+192

snowflake0531 · Answer

So add them together to get your answer

surjithayer · Answer

$$first~term=a _{1}=3$$
first we need to find the common ratio.
$$common~ratio~r=\frac{ a _{n} }{ a_{n-1} }=\frac{ any~term }{ previous ~term}=\frac{ a_{2} }{ a_{1}}=\frac{ -6 }{ 3}=-2   $$
here we took 2nd and first term.
$$S_{n}=a_{1}\frac{ 1-r^n }{ 1-r } (for~G.P.)$$
here n=7,plug the values and find the sum of first 7 terms.