If a_(1)=3 and a_(n)=4a_(n-1)-5 then find the value of a_(4). Answer:

Given terms: We are given the first term of the sequence, a_(1)=3, and the recursive formula a_(n)=4a_(n-1)-5. To find a_(4), we need to find the values of a_(2), a_(3), and then a_(4) using the recursive formula.Find a_(2): First, let's find a_(2) using the recursive formula: a_(2) = 4a_(1) - 5 a_(2) = 4(3) - 5 a_(2) = 12 - 5 a_(2) = 7 We have found that a_(2) is 7.Find a_(3): Next, we find a_(3) using the recursive formula and the value of a_(2): a_(3) = 4a_(2) - 5 a_(3) = 4(7) - 5 a_(3) = 28 - 5 a_(3) = 23 We have found that a_(3) is 23.Find a_(4): Finally, we find a_(4) using the recursive formula and the value of a_(3): a_(4) = 4a_(3) - 5 a_(4) = 4(23) - 5 a_(4) = 92 - 5 a_(4) = 87 We have found that a_(4) is 87.

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Find the roots of factored polynomials

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Q. If

a_{1}=3

and

a_{n}=4 a_{n-1}-5

then find the value of

a_{4}

\newline

Answer:

Given terms: We are given the first term of the sequence, $a_{1}=3$ , and the recursive formula $a_{n}=4a_{n-1}-5$ . To find $a_{4}$ , we need to find the values of $a_{2}$ , $a_{3}$ , and then $a_{4}$ using the recursive formula.
Find $a_{2}$ : First, let's find $a_{2}$ using the recursive formula: $\newline$ $a_{2} = 4a_{1} - 5$ $\newline$ $a_{2} = 4(3) - 5$ $\newline$ $a_{2} = 12 - 5$ $\newline$ $a_{2} = 7$ $\newline$ We have found that $a_{2}$ is $7$ .
Find $a_{3}$ : Next, we find $a_{3}$ using the recursive formula and the value of $a_{2}$ : $a_{3} = 4a_{2} - 5$ $a_{3} = 4(7) - 5$ $a_{3} = 28 - 5$ $a_{3} = 23$ We have found that $a_{3}$ is $23$ .
Find $a_{4}$ : Finally, we find $a_{4}$ using the recursive formula and the value of $a_{3}$ : $a_{4} = 4a_{3} - 5$ $a_{4} = 4(23) - 5$ $a_{4} = 92 - 5$ $a_{4} = 87$ We have found that $a_{4}$ is $87$ .