If a_(1)=7 and a_(n)=-2a_(n-1)-2 then find the value of a_(3). Answer:

Find a_(2): We are given the first term of the sequence, a_(1)=7, and the recursive formula for the sequence, a_(n)=-2a_(n-1)-2. To find a_(3), we first need to find a_(2) using the recursive formula. a_(2) = -2a_(1) - 2 = -2(7) - 2 = -14 - 2 = -16Find a_(3): Now that we have a_(2) = -16, we can use the recursive formula again to find a_(3). a_(3) = -2a_(2) - 2 = -2(-16) - 2 = 32 - 2 = 30

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If
a_(1)=7 and
a_(n)=-2a_(n-1)-2 then find the value of
a_(3).
Answer:

If $a_{1}=7$ and $a_{n}=-2 a_{n-1}-2$ then find the value of $a_{3}$ . $\newline$ Answer:

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Precalculus

Find trigonometric ratios using multiple identities

Full solution

Q. If

a_{1}=7

and

a_{n}=-2 a_{n-1}-2

then find the value of

a_{3}

\newline

Answer:

Find $a_{2}$ : We are given the first term of the sequence, $a_{1}=7$ , and the recursive formula for the sequence, $a_{n}=-2a_{n-1}-2$ . To find $a_{3}$ , we first need to find $a_{2}$ using the recursive formula. $\newline$ $a_{2} = -2a_{1} - 2$ $\newline$ $= -2(7) - 2$ $\newline$ $= -14 - 2$ $\newline$ $= -16$
Find $a_{3}$ : Now that we have $a_{2} = -16$ , we can use the recursive formula again to find $a_{3}$ . $\newline$ $a_{3} = -2a_{2} - 2$ $\newline$ $= -2(-16) - 2$ $\newline$ $= 32 - 2$ $\newline$ $= 30$