Question

What is the sum of the first 6 terms in this series?
6 + (-18) + 54 + (-162) ...
-1,092
-2,184
1,092
2,184

Answer 1

EXPLANATION

$\mathrm{A\: geometric\: sequence\: has\: a\: constant\: ratio\: }r\mathrm{\: and\: is\: defined\: by}\: a_n=a_1\cdot r^{n-1}$

Compute the ratios of all the adjacent terms:

$r=\frac{a_{n+1}}{a_n}$$\frac{-18}{6}=-3,\: \quad \frac{54}{-18}=-3,\: \quad \frac{-162}{54}=-3$$\mathrm{The\: ratio\: of\: all\: the\: adjacent\: terms\: is\: the\: same\: and\: equal\: to}$$r=-3$$a_n=a_1r^{\mleft\{n-1\mright\}}$$\mathrm{For\: }a_1=6,\: r=-3$$a_n=6\mleft(-3\mright)^{n-1}$

Now, plugging in the term 5 into the equation:

$a_n=6\cdot(-3)^{5-1}$

Subtracting numbers:

$a_n=6\cdot(-3)^4=6\cdot81=486$

The 5th term is 486.

Now, we need to compute the 6th term:

$a_n=6\cdot(-3)^{6-1}=6\cdot(-3)^5$

Computing the numbers:

$a_n=6\cdot(-243)$

Multiplying the numbers:

$a_n=-1,458$

The term 6th is -1458

Adding the first 6 terms:

6 + (-18) + 54 + (-162) + 486 + (-1,458) = -1,092

In conclusion, the solution is -1,092