Show that the following sequences are H.P. Also find 10th term .
$\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, . . .$

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Solution

Given series is in H.P.
$\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, . . . . . . . . .$

The given series in A.P becomes,
$2, 5, 8, 11, . . . . . . . . .$

So, $d = 5 - 2 = 8 - 5 = 3$

As common difference remains the same, given series is in A.P,

Implies that given series is in H.P
$a_{10} = a + (10 - 1) d$
$= 2 + 9 \times 3 = 29$

$∴ h_{10} = \frac{1}{29}$

Hence, this is the answer.

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Show that the following sequences are H.P. Also find 10th term .12,15,18,111,...

Given series is in H.P.12,15,18,111,.........The given series in A.P becomes,2,5,8,11,.........So, d=5−2=8−5=3As common difference remains the same, given series is in A.P,Implies that given series is in H.Pa10=a+(10−1)d=2+9×3=29∴h10=129Hence, this is the answer.

Show that the following sequences are H.P. Also find 10th term .
$\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, . . .$