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Given T = R + RS Lateral inversion makes it to be R + RS = T Taking R as common factor, we get R(1+S) = T Now dividing by (1+S) both sides, R = T / (1+S) Hence the solution R = T/(1+S)
r=0,Tr-r = 0 = r(T-1), since T != 1, then T-1 is non zero so r must be zero.
To find any term of a geometric sequence from another one you need the common ration between terms: t{n} = t{n-1} × r = t{1} × r^(n-1) where t{1} is the first term and n is the required term. It depends what was given in the geometric sequence ABOVE which you have not provided us. I suspect that along with the 10th term, some other term (t{k}) was given; in this case the common difference can be found: t{10} = 1536 = t{1} × r^9 t{k} = t{1} × r^(k-2) → t{10} ÷ t{k} = (t{1} × r^9) ÷ (t{1} × r^(k-1)) → t{10} ÷ t{k} = r^(10-k) → r = (t{10} ÷ t{k})^(1/(10-k)) Plugging in the values of t{10} (=1536), t{k} and {k} (the other given term (t{k}) and its term number (k) will give you the common ratio, from which you can then calculate the 11th term: t{11} = t(1) × r^9 = t{10} × r
D = rt t = d/r r = d/t
P V = n R TDivide each side by ( n T ):(P V) / (n T) = R