In a geometric progress, the third term is 16 and the fifth term is 4 .Calculate (i) The first term and the common ratio, (ii) The tenth term (iii) The sum to infinity

To solve this problem, let's denote the first term as \( a \) and the common ratio as \( r \). From the information given, we can formulate two equations: 1. The third term is given by \( ar^2 = 16 \) 2. The fifth term is given by \( ar^4 = 4 \) Now, we can set up the equations as follows: From the second equation, we can express \( a \) in terms of \( r \): \[ ar^4 = 4 \Rightarrow a = \frac{4}{r^4} \] Substituting this into the first equation: \[ \frac{4}{r^4} r^2 = 16 \] \[ \frac{4}{r^2} = 16 \] \[ 4 = 16r^2 \Rightarrow r^2 = \frac{1}{4} \Rightarrow r = \frac{1}{2} \text{ (taking the positive root for geometric progress)}\] Now, substituting \( r \) back to find \( a \): \[ a = \frac{4}{(\frac{1}{2})^4} = \frac{4}{\frac{1}{16}} = 64 \] Now we have: (i) The first term \( a = 64 \) and the common ratio \( r = \frac{1}{2} \). Next, let's find the tenth term: \[ T_{10} = ar^9 = 64 \left(\frac{1}{2}\right)^9 = 64 \cdot \frac{1}{512} = \frac{64}{512} = \frac{1}{8} \] (ii) The tenth term is \( \frac{1}{8} \). Finally, to find the sum to infinity of a geometric series where \( |r| < 1 \): \[ S_{\infty} = \frac{a}{1 - r} = \frac{64}{1 - \frac{1}{2}} = \frac{64}{\frac{1}{2}} = 128 \] (iii) The sum to infinity is \( 128 \).