If a_(1)=2 and a_(n+1)=-2a_(n)-3 then find the value of a_(4). Answer:

Given terms: We are given the first term of the sequence, a_(1)=2, and the recursive formula for the sequence, a_(n+1)=-2a_(n)-3. To find a_(4), we need to find a_(2), a_(3), and then a_(4) using the recursive formula.Find a_(2): Let's find a_(2) using the recursive formula. We substitute n=1 into the formula to get a_(2)=-2a_(1)-3. a_(2) = -2 * a_(1) - 3 a_(2) = -2 * 2 - 3 a_(2) = -4 - 3 a_(2) = -7Find a_(3): Now let's find a_(3) using the recursive formula. We substitute n=2 into the formula to get a_(3)=-2a_(2)-3. a_(3) = -2 * a_(2) - 3 a_(3) = -2 * (-7) - 3 a_(3) = 14 - 3 a_(3) = 11Find a_(4): Finally, let's find a_(4) using the recursive formula. We substitute n=3 into the formula to get a_(4)=-2a_(3)-3. a_(4) = -2 * a_(3) - 3 a_(4) = -2 * 11 - 3 a_(4) = -22 - 3 a_(4) = -25

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Q. If

a_{1}=2

and

a_{n+1}=-2 a_{n}-3

then find the value of

a_{4}

\newline

Answer:

Given terms: We are given the first term of the sequence, $a_{1}=2$ , and the recursive formula for the sequence, $a_{n+1}=-2a_{n}-3$ . To find $a_{4}$ , we need to find $a_{2}$ , $a_{3}$ , and then $a_{4}$ using the recursive formula.
Find $a_{2}$ : Let's find $a_{2}$ using the recursive formula. We substitute $n=1$ into the formula to get $a_{2}=-2a_{1}-3$ . $\newline$ $a_{2} = -2 \times a_{1} - 3$ $\newline$ $a_{2} = -2 \times 2 - 3$ $\newline$ $a_{2} = -4 - 3$ $\newline$ $a_{2} = -7$
Find $a_{3}$ : Now let's find $a_{3}$ using the recursive formula. We substitute $n=2$ into the formula to get $a_{3}=-2a_{2}-3.$ $\newline$ $a_{3} = -2 \times a_{2} - 3$ $\newline$ $a_{3} = -2 \times (-7) - 3$ $\newline$ $a_{3} = 14 - 3$ $\newline$ $a_{3} = 11$
Find $a_{4}$ : Finally, let's find $a_{4}$ using the recursive formula. We substitute $n=3$ into the formula to get $a_{4}=-2a_{3}-3$ .
$a_{4} = -2 \times a_{3} - 3$
$a_{4} = -2 \times 11 - 3$
$a_{4} = -22 - 3$
$a_{4} = -25$