Question

Find the 10th term of the sequence:
8, 8/3, 8/5............!?​

Answer 1

Answer:

Given sequence is in H.P

1/8 , 3/8 , 5/8 ...... is in AP with first terms a = 1 / 8.

Common difference = d = 1 / 4

now nth term of H.P = 1 / nth term of A.P

In (H.P) = 1 / Tn (A.P) = 1 / 1 / 8 + n - 4 / 4 = 8 / Tn - 1

T10 = 8 / 2 × 10 - 1 = 8 / 79

Answer 2

Given:

The sequence → 8, 8/3, 8/5..........

To Find:

${10}^{th} \ term$

Solution:

The sequence goes,

$8 \: .. \frac{8}{3} .. \frac{8}{5} ..$

• We can see that the numerator in each term remains the same (8 every time) however, the denominator changes.

• An internal sequence is formed within the denominators in these numbers.

• The internal sequence looks like

1.. 3.. 5..

The common difference (d) = 5 - 3 = 2

The first term of the sequence (a) = 1

10th term = ?

To find the nth term we must recall this formula

${n}^{th} \: term = a + (n - 1)d$

For 10th term:

Given,

a = 1

d = 2

n = 10

Put the given values in the formula to find the 10th term

${10}^{th} \: term = 1 + (10 - 1)2$

${10}^{th} \: term = 1 + 18$

${10}^{th} \: term = 19$

• Now, we know the 10th term of the internal sequence is 19. The 10th term of original sequence will have 19 as its denominator and 8 is numerator.

Hence, the 10th term of the sequence is

$\frac{8}{19}$