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

If $a_{1}=3$ and $a_{n+1}=3 a_{n}+5$ then find the value of $a_{4}$ . $\newline$ Answer:

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Find trigonometric ratios using multiple identities

Full solution

Q. If

a_{1}=3

and

a_{n+1}=3 a_{n}+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 for the sequence, $a_{n+1} = 3a_{n} + 5$ . To find $a_{4}$ , we need to find $a_{2}$ , $a_{3}$ , and then $a_{4}$ using the recursive formula.
Find $a_{2}$ : First, let's find $a_{2}$ using the recursive formula: $\newline$ $a_{2} = 3a_{1} + 5$ $\newline$ $a_{2} = 3 \times 3 + 5$ $\newline$ $a_{2} = 9 + 5$ $\newline$ $a_{2} = 14$
Find $a_{3}$ : Next, we'll find $a_{3}$ using the recursive formula and the value of $a_{2}$ we just found: $\newline$ $a_{3} = 3a_{2} + 5$ $\newline$ $a_{3} = 3 \times 14 + 5$ $\newline$ $a_{3} = 42 + 5$ $\newline$ $a_{3} = 47$
Find $a_{4}$ : Finally, we'll find $a_{4}$ using the recursive formula and the value of $a_{3}$ : $a_{4} = 3a_{3} + 5$ $a_{4} = 3 \times 47 + 5$ $a_{4} = 141 + 5$ $a_{4} = 146$

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