Calculate the least number of terms of the geometric progression 5 + 10 + 20 + ... whose sum would exceed 10,00,000.

Answer

To find the least number of terms of the geometric progression (GP) 5, 10, 20, ... whose sum exceeds 10,00,000, we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term \( a \) of the GP is 5, and the common ratio \( r \) is 2 (since \( 10/5 = 2 \) and \( 20/10 = 2 \)).

### Step 2: Write the formula for the sum of the first \( n \) terms of a GP
The sum \( S_n \) of the first \( n \) terms of a geometric progression is given by the formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Substituting the values of \( a \) and \( r \):
\[ S_n = \frac{5(2^n - 1)}{2 - 1} = 5(2^n - 1) = 5 \cdot 2^n - 5 \]

### Step 3: Set up the inequality for the sum
We need to find the smallest \( n \) such that:
\[ S_n > 10,00,000 \]
This gives us the inequality:
\[ 5 \cdot 2^n - 5 > 10,00,000 \]

### Step 4: Simplify the inequality
Adding 5 to both sides:
\[ 5 \cdot 2^n > 10,00,005 \]
Dividing both sides by 5:
\[ 2^n > 2,00,001 \]

### Step 5: Solve for \( n \)
To find \( n \), we can take the logarithm base 2 of both sides:
\[ n > \log_2(2,00,001) \]

### Step 6: Calculate \( \log_2(2,00,001) \)
Using the change of base formula:
\[ \log_2(2,00,001) = \frac{\log_{10}(2,00,001)}{\log_{10}(2)} \]
Calculating \( \log_{10}(2,00,001) \) and \( \log_{10}(2) \):
- \( \log_{10}(2) \approx 0.301 \)
- \( \log_{10}(2,00,001) \approx 5.301 \)

Now substituting:
\[ n > \frac{5.301}{0.301} \approx 17.6 \]

Since \( n \) must be a whole number, we round up to the next integer:
\[ n \geq 18 \]

### Conclusion
The least number of terms required in the geometric progression such that their sum exceeds 10,00,000 is \( n = 18 \).

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