The sum of the first seven terms of a geometric sequence is 127 and the quotient is 2, find the terms of the sequence.

Tuesday, April 26, 2016

The sum of the first seven terms of a geometric sequence is 127 and the quotient is 2, find the terms of the sequence.

Hello!

Probably we need to find all the terms of this progression.

Recall that each next term of a geometric progression is obtained by multiplying the previous term by the quotient, denote it as $\displaystyle{q}.$ Denote the first term as $\displaystyle{b},$ then the second is $\displaystyle{b}\cdot{q},$ the third is $\displaystyle{b}\cdot{q}\cdot{q}={b}\cdot{{q}}^{{2}},$ and the k-th term is $\displaystyle{b}\cdot{{q}}^{{{k}-{1}}}.$

It is also well-known that the sum of $\displaystyle{n}$ terms of a geometric progression with the first term $\displaystyle{b}$ and $\displaystyle{q}\ne{1}$ is $\displaystyle{S}={b}\cdot\frac{{{{q}}^{{n}}-{1}}}{{{q}-{1}}}.$

In our case $\displaystyle{q}={2},$ $\displaystyle{b}$ is unknown, $\displaystyle{n}={7}$ and $\displaystyle{S}={127}.$ This way we obtain a simple equation for $\displaystyle{b}:$

$\displaystyle{127}={b}\cdot\frac{{{{2}}^{{7}}-{1}}}{{{2}-{1}}}={b}\cdot{127},$ because $\displaystyle{{2}}^{{7}}={128}$ (check by multiplying 2*2*2*2*2*2*2).

Obviously the only solution is $\displaystyle{b}={1},$ and the entire array is 1, 2, 4, 8, 16, 32, 64.

We could solve this problem without the formula for sum, adding all $\displaystyle{7}$ terms manually:

$\displaystyle{b}+{2}{b}+{4}{b}+{8}{b}+{16}{b}+{32}{b}+{64}{b}={127},$ the left side is $\displaystyle{\left({1}+{2}+{4}+{8}+{16}+{32}+{64}\right)}{b}={127}{b}={127},$ and again $\displaystyle{b}={1}.$

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Tuesday, April 26, 2016