Answer:
[tex] \small{\boxed{ \sf{Term \: in \: both \: the \: sequence \: is \: 15}}}[/tex]
Step-by-step explanation:
To determine the common term in both the sequence let's solve for each sequence individually.
Let's solve for first sequence
[tex]f(n) = f(n-1)+4[/tex]
we are already told that f(1) = 3
let's solve for f(2),f(3),f(4)........ and so on till we get the highest digit or more than the highest digit from given option.
f(2) = f(2-1)+4 = f(1)+4
f(2) = 3+4
[tex]\small{ \fbox{f(2) = 7}}[/tex]
f(3) = f(3-1)+4 = f(2)+4
f(3) = 7+4
[tex] \small{\fbox{f(3)= 11}}[/tex]
on solving further..
The series obtained is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 111, 115, 119, 123, 127, 131, 135.....
Let's solve for second series
[tex]f(n) = 3×5^{n -1}[/tex]
given that f(1) = 3,
f(2) = 3×5¹ = 15
f(3) = 3×5² = 75
f(4) = 3×5³ = 375
The series obtained is 3, 15, 75, 375.....
The term obtained in both the sequence is 15 hence option B is correct Answer.
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