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

Question

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

UpStudy AI · Accepted Answer

To solve the problem, we will use the properties of a geometric progression (GP).

In a GP, the $$ n $$-th term can be expressed as:
$$
T_n = a \cdot r^{n-1}
$$
where:
- $$ T_n $$ is the $$ n $$-th term,
- $$ a $$ is the first term,
- $$ r $$ is the common ratio,
- $$ n $$ is the term number.

### Given:
- The third term $$ T_3 = 16 $$
- The fifth term $$ T_5 = 4 $$

### Step 1: Set up the equations
From the definitions of the terms, we can write:
1. $$ T_3 = a \cdot r^{2} = 16 $$  (Equation 1)
2. $$ T_5 = a \cdot r^{4} = 4 $$   (Equation 2)

### Step 2: Solve the equations
We can express $$ a $$ from Equation 1:
$$
a = \frac{16}{r^2}
$$
Now, substitute $$ a $$ into Equation 2:
$$
\frac{16}{r^2} \cdot r^4 = 4
$$
This simplifies to:
$$
16r^2 = 4
$$
Dividing both sides by 4:
$$
4r^2 = 1
$$
Thus:
$$
r^2 = \frac{1}{4}
$$
Taking the square root:
$$
r = \frac{1}{2} \quad \text{(since the common ratio is positive)}
$$

Now substitute $$ r $$ back into the expression for $$ a $$:
$$
a = \frac{16}{\left(\frac{1}{2}\right)^2} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = 64
$$

### Step 3: Calculate the tenth term
The tenth term $$ T_{10} $$ is given by:
$$
T_{10} = a \cdot r^{9}
$$
Substituting the values of $$ a $$ and $$ r $$:
$$
T_{10} = 64 \cdot \left(\frac{1}{2}\right)^{9} = 64 \cdot \frac{1}{512} = \frac{64}{512} = \frac{1}{8}
$$

### Step 4: Calculate the sum to infinity
The sum to infinity $$ S $$ of a geometric series is given by:
$$
S = \frac{a}{1 - r}
$$
Substituting the values of $$ a $$ and $$ r $$:
$$
S = \frac{64}{1 - \frac{1}{2}} = \frac{64}{\frac{1}{2}} = 64 \cdot 2 = 128
$$

### Summary of Results
(i) The first term $$ a = 64 $$ and the common ratio $$ r = \frac{1}{2} $$  
(ii) The tenth term $$ T_{10} = \frac{1}{8} $$  
(iii) The sum to infinity $$ S = 128 $$

Thus, the final answers are:
- First term: $$ 64 $$
- Common ratio: $$ \frac{1}{2} $$
- Tenth term: $$ \frac{1}{8} $$
- Sum to infinity: $$ 128 $$
First term: 64 Common ratio: 1/2 Tenth term: 1/8 Sum to infinity: 128

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

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