In the following sequence of numbers, each term, starting with the third, is obtained by adding three times the previous term plus twice the previous term to the previous one: $$a_1, a_2, 78, a_4, a_5, 3438, a_7, a_8,…$$As seen in the sequence, the third term is $78$ and the sixth term is $3438$. What is the value of the term $a_7$?
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Tags: algebra
miyukina
26.03.2024 08:57
78 = 2a + 3b T(4) = 2b + 3(2a + 3b) = 6a + 11b T(5) = 2(2a + 3b) + 3(6a + 11b) = 22a + 39b T(6) = 78a + 139b = 3438 T(7) = 278a + 495b 78a + 139b = 3438 – (2a + 3b = 78) × 39 22b = 3438 – 3042 ===> b = 396/22 = 18 ===> a = 12 T(7) = 12246
BackToSchool
26.03.2024 09:05
$$\begin{cases}
a_1 &= a_1 \\
a_2 &= a_2 \\
a_3 &= 3a_2 + 2a_1 =78\\
a_4 &= 3a_3 + 2a_2 = 234+2a_2\\
a_5 &= 3a_4 + 2a_3 = 858+ 6a_2\\
a_6 &= 3a_5 + 2a_4 = 3042 + 22a_2 =3438\\
\end{cases}
\implies a_2 = 18, a_1 = 12$$$$\implies a_7= 3a_6 + 2a_5 = 3 \times 3438 + 2 \times (858 + 6a_2) = \boxed {12246}$$