If a_(1)=7 and a_(n+1)=-3a_(n)+5 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+1)=-3a_(n)+5. To find a_(3), we first need to find a_(2) using the recursive formula. Substitute n=1 into the recursive formula to find a_(2): a_(2) = -3a_(1) + 5 a_(2) = -3(7) + 5 a_(2) = -21 + 5 a_(2) = -16Find a_(3): Now that we have a_(2)=-16, we can use the recursive formula again to find a_(3). Substitute n=2 into the recursive formula to find a_(3): a_(3) = -3a_(2) + 5 a_(3) = -3(-16) + 5 a_(3) = 48 + 5 a_(3) = 53

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If
a_(1)=7 and
a_(n+1)=-3a_(n)+5 then find the value of
a_(3).
Answer:

If $a_{1}=7$ and $a_{n+1}=-3 a_{n}+5$ then find the value of $a_{3}$ . $\newline$ Answer:

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Precalculus

Find trigonometric ratios using multiple identities

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Q. If

a_{1}=7

and

a_{n+1}=-3 a_{n}+5

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, $a_{n+1}=-3a_{n}+5$ . To find $a_{3}$ , we first need to find $a_{2}$ using the recursive formula. $\newline$ Substitute $n=1$ into the recursive formula to find $a_{2}$ : $\newline$ $a_{2} = -3a_{1} + 5$ $\newline$ $a_{2} = -3(7) + 5$ $\newline$ $a_{2} = -21 + 5$ $\newline$ $a_{1}=7$ $0$
Find $a_{3}$ : Now that we have $a_{2}=-16$ , we can use the recursive formula again to find $a_{3}$ . Substitute $n=2$ into the recursive formula to find $a_{3}$ : $a_{3} = -3a_{2} + 5$ $a_{3} = -3(-16) + 5$ $a_{3} = 48 + 5$ $a_{3} = 53$