# If the first two terms of H.P. are 2/5 and 12/13 find the largest term of the H.P.

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2/5 , 12/13, ....

a1= 5/2

a2= 13/12

Let us find the ratio between terms:

r= 13/12 - 5/2

= (13-30)/12= -17/12

Then:

an = a1+ (n-1)*(-17/12)

= (5/2) + 17/12 - 17n/12

= (30+17 - 17n)/12

==> an = (47-17n)/12

let us find the smallest of an

The smallest of an is the smallest of (47-17n)

but n = 1,2,3,....

==> the smallest of 47-17n is when n largest

==> n = 2 is the largest

==> a2 is the smallest

==> but an= 13/12

==> 1/an is the largest

==> 12/13 is the largest

We'll note the first 3 terms of H.P. as:

1/a , 1/(a+d) , 1/(a+2d) , .............

The first term of H.P. is t1 = 1/a

But, from enunciation, t1 = 2/5

So, 2/5 = 1/a

We'll cross multiply => 2a = 5

We'll divide by 2:

a = 5/2

Also, from enunciation, we'll have:

t2 = 12/13

But a2 = a1 + d = a + d

So, d = a2 - a1

d = 13/12 - 5/2

d = (13-30)/12

d = -17/12

tn = 1/[a+(n-1)d]

tn = 1/[5/2 + (n-1)(-17/12)]

tn = 12/(30-17n+17)

tn = 12/(47-17n)

tn is the largest term, when the denominator is least.

For n = 2 => 47-17*2 = 47 - 34 = 13 => tn = 12/13

**So, the largest term of H.P. is 12/13.**