Statement 1 `1,2,4,8,"….."` is a GP,`4,8,16,32,"…."` is a GP and `1+4,2+8,4+16,8+32,"…."` is also a GP. Statement 2 Let general term of a GP with common ratio r be `T_(k+1)` and general term of another GP with common ratio r be `T'_(k+1)`, then the series whode general term `T''_(k+1)=T_(k+1)+T'_(k+1)` is also a GP woth common ratio r.

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a corrct explanation for Statement 1.
B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
C. Statement 1 is true, Statement 2 is false.
D. Statement 1 is false, Statement 2 is true.

Explanation

To solve the problem step by step, we will analyze both statements and verify their validity.

### Step 1: Analyze Statement 1
We have two sequences:
1. \(1, 2, 4, 8, \ldots\)
2. \(4, 8, 16, 32, \ldots\)

**First Sequence:**
- This sequence is a geometric progression (GP) with the first term \(a = 1\) and common ratio \(r = 2\).
- The general term of this GP can be expressed as:
\[ T_{k+1} = a \cdot r^{k} = 1 \cdot 2^{k} = 2^{k} \]

**Second Sequence:**
- This sequence is also a GP with the first term \(a = 4\) and common ratio \(r = 2\).
- The general term of this GP can be expressed as:
\[ T'_{k+1} = 4 \cdot 2^{k} = 4 \cdot 2^{k} \]

### Step 2: Analyze the Combined Sequence
Now, we need to analyze the sequence formed by the sums:
- \(1 + 4, 2 + 8, 4 + 16, 8 + 32, \ldots\)

This can be expressed as:
\[ T''_{k+1} = T_{k+1} + T'_{k+1} \]
Substituting the general terms we found:
\[ T''_{k+1} = 2^{k} + 4 \cdot 2^{k} = (1 + 4) \cdot 2^{k} = 5 \cdot 2^{k} \]

### Step 3: Determine if the Combined Sequence is a GP
- The general term \(T''_{k+1} = 5 \cdot 2^{k}\) is also a geometric progression with the first term \(5\) and common ratio \(2\).

### Conclusion for Statement 1
Since both sequences are GPs and their sum also forms a GP, Statement 1 is **True**.

### Step 4: Analyze Statement 2
Statement 2 claims that if we have two GPs with common ratio \(r\), then the series formed by their sums is also a GP with the same common ratio.

This is indeed true as shown in the analysis above:
- If \(T_{k+1} = a \cdot r^{k}\) and \(T'_{k+1} = b \cdot r^{k}\), then:
\[ T''_{k+1} = T_{k+1} + T'_{k+1} = (a + b) \cdot r^{k} \]
This shows that \(T''_{k+1}\) is also a GP with common ratio \(r\).

### Conclusion for Statement 2
Statement 2 is also **True** and serves as a correct explanation for Statement 1.

### Final Answer
Both statements are true, and Statement 2 correctly explains Statement 1. Therefore, the answer is **A**.

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