# Find the 10th and nth terms of the sequence 4,30/7,60/13,...

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4/1, 30/7 , 60/13, .....

Let us write as follow:

1/4, 7/30, 13/60

Now let us calculate the ratio between terms.

r= 7/30 - 1/4 = 13/60 - 7/30

= (28-30)/120 = 13-14/60

r = -1/60 = -1/60

Then:

a1= 1/4

a2= 1/4 + -1/60

......

a10 = 1/4 + (9*-1/60)

= 1/4 - 9/60

= 1/4 - 3/20

= 5-3/20 = 2/20 = 1/10

==> a10 = 1/10

==> an = 1/4 + (n-1)*-1/60

= 1/4 + 1/60 - n/60

= (15 + 1 - n)/60

==? an = (16-n)/60

If we'll write the sequence of reciprocals:

1/4, 7/30, 13/60, ....

and we'll calculate the difference between consecutive terms:

7/30 - 1/4 = -1/60

13/60 - 7/30 = -1/60

..................................

we'll notice that we'll obtain the common difference d = -1/60, so the sequence of reciprocal terms is an a.s.

So, the given sequence is a harmonic sequence.

Now, we'll calculate a10 for the arithmetic sequence:

a10 = a1 + 9d

a10 = 1/4 + 9*(-1/60)

a10 = 1/4 - 3/20

a10 = (5-3)/20

a10 = 1/10

**So, the 10-th term of the H.P. is t10 = 10.**

The n-th of the A.P. is:

an = a1 + (n-1)d

an = 1/4 + (n-1)(-1/60)

an = (15 - n + 1)/60

an = (16-n)/60

**The n-th term of the H.P. is tn = 60/(16-n).**