The first four term of sequeces A,B,C,D are shown Sequence A {(1/3),(2/4),(3/5),(4/6)}

Question

The first four term of sequeces A,B,C,D are shown
Sequence A

(1 / 3), (2 / 4), (3 / 5), (4 / 6)

Answer 1

Step 1:
We know that a sequence isa function from nautral number to real number.
Let,(: $a_{n}$ :)be a sequence the required sequence,
Then,
$(: a_{n} :) = a_{1}, a_{2}, a_{3}, a_{4} . . . . . a_{n}, . . .$
Given
$a_{1} = (1 / 3)$
$a_{2} = (2 / 4)$
$a_{3} + (3 / 5)$
$a_{4} (4 / 6)$
$a_{5} = (5 / 7)$
then, the $n^{t h}$ sequence (: $a_{n}$ :) is given by
$a_{n} = \frac{n}{n + 2}$
now verify
for $n = 1, a_{1} = \frac{1}{1 + 2} = \frac{1}{3}$
for $n = 2, a_{2} = \frac{2}{2 + 2} = \frac{2}{4}$
for $n = 3, a_{3} = \frac{3}{3 + 2} = \frac{3}{5}$
for $n = 4, a_{4} = \frac{4}{4 + 2} = \frac{4}{6}$

The first four term of sequeces A,B,C,D are shown Sequence A {(1/3),(2/4),(3/5),(4/6)}

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