Osoayobami: THE MILITARY IN WEST AFRICA ?Nigerian,ghana,sierra leone all retained military forces at the time of their independence.these armies had all grown out of the colonial royal west african frontier forces(RWAFF).THE RWAFF Had earlier been set up to protect the interest of the british within the west african empire.the army was enlarged at different times to suppress revolt and was essentially a tool for repression.the gambia which is a small country did not have a standing army,but instead had a small field-force of about 50
men.these were like armed policemen.in order to guarantee the loyalty of the military and to make sure units did not go against the interest of the colonial gorverment,africans were kept to the low ranks.most of the officer were british.

Alpha Maximus: alright, here we go:
Let the first term be 'a' and the common difference be 'x'
1st term : a
2nd term : a+x
3rd term : a + 2x
Since the sum of the product of just two terms and 1 give the same answer, we have that:
(a(a+x))+1=(a(a+2x))+1=((a+x)(a+2x))+1
Therefore,
A^2 +ax+1=a^2+2ax+1=a^2+3ax+2x^2+1
Now there are three equations, we equate the first and last, and temporarily isolate the second(because we don't know which is equal to which yet)
a^2+3ax+2x^2+1=a^2+ax+1
Collecting like terms we have
2ax+2x^2=0
2x(a+x)=0
X=0(impossible,since its an arithmetic progression,there has to be a difference) and a+x=0, a=-x(correct)
Since 2ax+2x^2=0, the middle equation is also equal to 0
Thus,
a^2+2ax+1=0
Now substitute 'a=-x' into the middle/second equation
(-x)^2+2(-x)x+1=0
X^2-2x^2+1=0
Factorising, we have:
(X-1)(x-1)=0
X=1(twice)
Now that we have established the common difference of '1', we input it back into the original equations but this time we subtract 1 from all three sides for simplicity and we have:
a(a+1)=a(a+2)=(a+1)(a+2)
We first equate and solve the first and second
a(a+1)=a(a+2)
a^2+a=a^2+2a
Collecting like terms, we have
-a=0
Thus, a=0
Now we simply equate the third equation to 0
(a+1)(a+2)=0
Thus a=-1 and a=-2
We try out a=-2 as the first term of the series a, a+1, a+2 but it does not satisfy the question
But when a=-1 is tested, it satisfies the question(all products of a pair + 1=a perfect square)
Therefore the terms are -1, -1+1, -1+2=-1, 0 , 1
N.B: recall that in my solution , the middle equation was already equal to 0, before I arrived at my final three answers

benbuks: ..my dear, i actually dnt kw what to tell him again,so i quit responding to his post, imagine i wast'nt even referin 2him..bt paul....please lets let him be...lets continue solving., learning frm each oda.nd helpin doz dat r ready 2 learn....

benbuks: ..my dear, i actually dnt kw what to tell him again,so i quit responding to his post, imagine i wast'nt even referin 2him..bt paul....please lets let him be...lets continue solving., learning frm each oda.nd helpin doz dat r ready 2 learn....

doubleDx:

With this crap & pride you would never learn anything....Are you saying the way you applied your log law was right? Man, you are a fuckin math kid...I don't argue with folks like you! mtchwww. I don't know what you think you know in math now....I dey laugh in tongues! Na una type dey claim say una write exams well but dem give una bad grades...yeye!