Find the next five terms of each of the following sequences given by:
$a_{1} = - 1, a_{n} = \frac{a_{n - 1}}{n}, n \geq 2$

Open in App

Solution

The $n^{t h}$ term of the sequence is
$a_{n} = \frac{a_{n - 1}}{n} f o r n \geq 2 . . . . . . . . . . . . . e q (1)$
and $a_{n} = - 1 f o r n = 1$
$⟹ a_{1} = - 1 . . . . . . . e q (2)$

1) To find the second term of sequence $(a_{2})$ put n=2 in eq(1)
$⟹ a_{2} = \frac{a_{2 - 1}}{2}$

$⟹ a_{2} = \frac{a_{1}}{2}$
put value of $a_{1}$ from eq(2)
$⟹ a_{2} = \frac{- 1}{2} . . . . e q (3)$

2)To find the third term of seqence $(a_{3})$ put n=3 in eq(1)
$⟹ a_{3} = \frac{a_{3 - 1}}{3}$

$⟹ a_{3} = \frac{a_{2}}{3}$

put value of $a_{2}$ from eq(3)

$⟹ a_{3} = \frac{- 1}{2 \times 3}$

$⟹ a_{3} = \frac{- 1}{6} . . . . . . . e q (4)$

3)To find the fourth term of sequence $(a_{4})$ put n=4 in eq(1)
$⟹ a_{4} = \frac{a_{4 - 1}}{4}$

$⟹ a_{4} = \frac{a_{3}}{4}$
put value of $a_{3}$ from eq(4)
$⟹ a_{4} = \frac{- 1}{6 \times 4}$

$⟹ a_{4} = \frac{- 1}{24} . . . . . . . e q (5)$

4)To find the fifth term of sequence $(a_{5})$ put $n = 5$ in eq(1)
$⟹ a_{5} = \frac{a_{5 - 1}}{5}$

$⟹ a_{5} = \frac{a_{4}}{5}$
put value of $a_{4}$ from eq(5)
$⟹ a_{5} = \frac{- 1}{24 \times 5}$

$⟹ a_{5} = \frac{- 1}{120} . . . . . . . e q (6)$

5)To find the sixth term of sequence $(a_{6})$ put n=6 in eq(1)
$⟹ a_{6} = \frac{a_{6 - 1}}{6}$

$⟹ a_{6} = \frac{a_{5}}{6}$

put value of $a_{5}$ from eq(6)

$⟹ a_{6} = \frac{- 1}{120 \times 6}$

$⟹ a_{6} = \frac{- 1}{720} . . . . . . . e q (7)$

Suggest Corrections

Similar questions

a_{1} = a_{2} = 2, a_{n} = a_{n - 1} - 3, n > 2

a_{1} = 4, a_{n} = 4 a_{n - 1} + 3, n > 1

a_{1} = - 1, a_{n} = \frac{a_{n} - 1}{n}, n \geq 2

a_{1} = 2, a_{2} = 3 + a_{1}

a_{n} = 2 a_{n - 1} + 5

n > 2

a_{1} = - 1, a_{n} = \frac{a_{n - 1}}{n + 2}, n > 1

\forall n ϵ N

Join BYJU'S Learning Program