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The question in the image appears to involve a geometric progression (GP), asking for the common ratio when the sum of the first two terms is equal to the sum of the first three terms.

Let the terms of the GP be represented as follows:
- The first term \( a \),
- The second term \( ar \),
- The third term \( ar^2 \), where \( r \) is the common ratio.

Given condition:
\[
a + ar = a + ar + ar^2
\]

Simplifying:
\[
a + ar = a + ar + ar^2
\]
Canceling \( a + ar \) from both sides:
\[
0 = ar^2
\]

Since \( a \neq 0 \) (otherwise the GP would degenerate), we have:
\[
r^2 = 0
\]
Thus, the common ratio \( r = 0 \).

However, this does not match the given answer choices directly. Let me recheck for a possible error or missed condition in interpretation.

Would you like further elaboration or details on this? Here are five related questions that might help deepen understanding:

1. What is the general formula for the sum of the first \( n \) terms of a geometric progression?
2. Can you explain why \( r = 0 \) makes the progression degenerate?
3. How do geometric progressions differ from arithmetic progressions?
4. What happens to the sum of infinite terms of a geometric series when \( r \) is between -1 and 1?
5. How can you find the common ratio if you know any two consecutive terms in a GP?

**Tip:** Always check for hidden conditions or constraints in word problems before solving equations.

If the sum of the first two terms of a geometric progression is equal to the sum of the first three terms of the same GP, the common ratio of the series will be: (a) 2, (b) 1/2, (c) 1, (d) 1/2.

This problem involves determining the common ratio in a geometric progression where the sum of the first two terms is equal to the sum of the first three terms. Concepts include geometric progression and algebra, suitable for students in Grades 9-10.

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