INDETERMINATE. PROBLEMS, 
199 
must be united by twos, threes, or fours, &c. according 
as one and thesame fraction occurs every second, third, 
or fourth, Sec. term (the odd.terms, when there happen 
any over, being always to be set aside, at the begin 
ning of the series). And it may be observed farther, 
that, to determine the sum of the progression thus 
arising, it will he sufficient to find the first term only, 
by an actual addition; since, not only the number of 
terms, but the common difference also, will be known; 
being always equal to the common difference of the 
limits of z (or of the quotients in the said third column) 
multiplied by the square of the number of terms united 
into one; whereof, the reason is evident. But all this 
relates to the cases wherein the co-efficients of the inde 
terminate, quantities, in the given equations, are (two 
*©f them at least) prime to each other: I shall add one 
example more, to shew the way of proceeding when 
those co-efficients admit of a common measure. 
problem xv, 
Supposing I2.r l- 1.5// 4-202 rr 100001 ; it is required 
to find the number of all the answers in positive integers. 
It is evident, by transposing 20: and dividing by (3) 
the greatest common measure of ¿r and //, that 4r i- 5//, 
and consequently its equal 33333 — 62 
must'be an integer, and therefore 22 — 2 divisible by 
3: but 32 is divisible by 3, and so the difference of 
these two, which is 2 4- 2, must be likewise divisible 
by the same number, and consequently 2 22 1 some 
multiple of 3. Make, therefore, l 4- 3u 22 2 (u be 
ing an integer): then the given equation by Substituting 
this value, will become 12a’ i- 1.5// f- 60^ f 20 — 
100001 ; which, bv division, See. is re luced to 4x +- 
5// 4- 20u — 33327: wherein the co-efficients of x and 
y are now prime to each otlier, and we are to find the 
number of all the variations, answering to the different 
interpretations of u, from 0 to the greatest limit in 
clusive.