The sum of the first 10 terms of a GP is equal to the sum of the first 12…

Given:
Sum of the first 10 terms = Sum of the first 12 terms
⇒ S₁₂ − S₁₀ = 0
⇒ 11th term + 12th term = 0
Let the first term = a and the common ratio = r. Then:
11th term = ar¹⁰ and 12th term = ar¹¹
So, ar¹⁰ + ar¹¹ = 0
⇒ ar¹⁰(1 + r) = 0
The common ratio is non-zero (given), so r ≠ 0, and a ≠ 0 (otherwise every sum would be 0, contradicting S₁₇ = 42). Hence ar¹⁰ ≠ 0, which forces 1 + r = 0, so r = −1.
Now, with r = −1 the GP becomes a, −a, a, −a, …
Since 17 is odd, the terms pair off and cancel, leaving:
S₁₇ = a = 42
Fourth term: T₄ = ar³ = 42 × (−1)³ = −42.