Recent content by tug187

Ok... ifwe know that term 3 is -3, a3 = a1*r^3-1 -3 = a1*2^2 -3 = 4*a1 -3/4 = a1 should be right

64 = r6 ? 26 = r6 r = 2? anyone kno if this is right?

thats what i don't understand, on other questions i got the common ratio easily by using the term before it but these 2 terms r not after each other so i don't know how to figure them out to make it more clear its: -3=a1r2 -192=a1r8

thats what I am wondering.. I don't think the way your thinking of works.. I forgot how to figure out the common ratio that's all i really need to know

question - if the 3rd and 9th term of a geometric series with a positive common ratio are -3 and -192 respectively, determine the value fo the first term, a. I kno we using a_n=a_1r^n-1 From that i got this : -3 = a_1r^2 -192 = a_1r^8 But I don't kno how I can solve for r or...