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Which recursive sequence would produce the sequence 
8,37,153,dots ?

a_(1)=8 and 
a_(n)=4a_(n-1)+5

a_(1)=8 and 
a_(n)=5a_(n-1)+4

a_(1)=8 and 
a_(n)=-3a_(n-1)+5

a_(1)=8 and 
a_(n)=5a_(n-1)-3

Which recursive sequence would produce the sequence 8,37,153, 8,37,153, \ldots ?\newlinea1=8 a_{1}=8 and an=4an1+5 a_{n}=4 a_{n-1}+5 \newlinea1=8 a_{1}=8 and an=5an1+4 a_{n}=5 a_{n-1}+4 \newlinea1=8 a_{1}=8 and an=3an1+5 a_{n}=-3 a_{n-1}+5 \newlinea1=8 a_{1}=8 and an=5an13 a_{n}=5 a_{n-1}-3

Full solution

Q. Which recursive sequence would produce the sequence 8,37,153, 8,37,153, \ldots ?\newlinea1=8 a_{1}=8 and an=4an1+5 a_{n}=4 a_{n-1}+5 \newlinea1=8 a_{1}=8 and an=5an1+4 a_{n}=5 a_{n-1}+4 \newlinea1=8 a_{1}=8 and an=3an1+5 a_{n}=-3 a_{n-1}+5 \newlinea1=8 a_{1}=8 and an=5an13 a_{n}=5 a_{n-1}-3
  1. Check First Option: Let's start by checking the first option:\newlinea1=8a_{1}=8 and \newlinean=4an1+5a_{n}=4a_{n-1}+5\newlineWe know the first term a1a_{1} is 88. Let's find the second term using the given recursive formula.\newlinea2=4a1+5a_{2}=4a_{1}+5\newlinea2=4×8+5a_{2}=4\times 8+5\newlinea2=32+5a_{2}=32+5\newlinea2=37a_{2}=37
  2. Find Second Term: Now let's find the third term using the same formula.\newlinea3=4a2+5a_{3}=4a_{2}+5\newlinea3=4×37+5a_{3}=4\times37+5\newlinea3=148+5a_{3}=148+5\newlinea3=153a_{3}=153
  3. Find Third Term: Since the given sequence is 8,37,1538, 37, 153, and we have found that the recursive formula an=4an1+5a_{n}=4a_{n-1}+5 produces the same sequence for the first three terms, we can conclude that this is the correct recursive sequence.

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