# One afternoon Danny was counting the fishes in the aquarium of the nerby pet-shop. He has counted r big red ones s small striped ones altogether. He did not wxpose the results to his sister Cathy,...

One afternoon Danny was counting the fishes in the aquarium of the nerby pet-shop. He has counted r big red ones s small striped ones altogether. He did not wxpose the results to his sister Cathy, however, he provided her with the following information: "The numbers 4, r and s in this order are consecutive terms of a geometric progression while the numbers r , s and 40 in this order are consecutive terms of an arithmetic progression." Find the number of big red fish and also the number of samll striped fish. Danny has managed to count in the aquarium.

*print*Print*list*Cite

In the given problem, 4, r, s are the consecutive terms of a geometric progressionn So, they will have a common ratio.

Therefore,` r/4=s/r` ................(i)

Again, r, s, 40 are the consecutive terms of an arithmetic progression.

So, they will have a common difference.

Therefore, s-r=40-s

or, 2s=40+r

or, s=`(40+r)/2` ..................(ii)

Substituting the value of s in equation (i), we get:

`r/4=(40+r)/(2r)`

`rArr r^2=80+2r`

`rArr r^2-2r-80=0`

`rArr r^2-10r+8r-80=0`

`rArr r(r-10)+8(r-10)=0`

`rArr (r-10)(r+8)=0`

Since r denotes the no. of big red fishes, we will discard the negative value.

Hence, r=10

Therefore, s=(40+10)/2=50/2=25.

**Hence,the number of big red fish and the number of small striped fish that Danny has managed to count in the aquarium are 10 and 25.**