Question

Find the first 3 terms of the series
a(n) = (n-3)/4.​

Answer 1

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Solution :-

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$a(n) = \frac{n - 3}{4} \\ a(1) = \frac{1 - 3}{4} = \frac{ - 2}{4} = \frac{ - 1}{2} \\ a(2) = \frac{2 - 3}{4} = \frac{ - 1}{4} \\ a(3) = \frac{3 - 3}{4} = 0$

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Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$


To find the first 3 terms of the sequence,

$a_{n} = \frac{n - 3}{4}$


$\bf{When \: n = 1,}$

$a_{1} = \frac{1 - 3}{4}$

$\: \: \: \: \: = \frac{ - 2}{4}$

$\: \: \: \: \: = \frac{ - 1}{2}$


$\bf{When \: n = 2 ,}$

$a_{2} = \frac{2 - 3}{4}$

$\: \: \: \: \: = \frac{ - 1}{4}$


$\bf{when \: n = 3 ,}$

$a_{3} = \frac{3 - 3}{4}$

$\: \: \: \: \: = \frac{0}{4}$

$\: \: \: \: \: = 0$


Therefore, the first 3 terms of the sequence are :

$\boxed{ \bf{ \frac{ - 1}{ \: 2} , \: \frac{ - 1}{ \: 4} \: \: and \: \: 0}}$