Suppose x₁, x₂, x₃, …, x₁₀₀ are in arithmetic progression such that x₅ = -4 and 2x₆ + 2x₉ = x₁₁ + x₁₃. Then, x₁₀₀ equals
Question:
Suppose x₁, x₂, x₃, …, x₁₀₀ are in arithmetic progression such that x₅ = -4 and 2x₆ + 2x₉ = x₁₁ + x₁₃. Then, x₁₀₀ equals
Suppose x₁, x₂, x₃, …, x₁₀₀ are in arithmetic progression such that x₅ = -4 and 2x₆ + 2x₉ = x₁₁ + x₁₃. Then, x₁₀₀ equals
Options
Answer: -194
Explanation:
Given: x5 = a + 4d = −4 → a + 4d = −4. 2x6 + 2x9 = x11 + x13. Compute terms: x6 = a + 5d x9 = a + 8d x11 = a + 10d x13 = a + 12d Plug into (2): 2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d) → 2a + 10d + 2a + 16d = 2a + 22d → 4a + 26d = 2a + 22d → 2a + 4d = 0 → a + 2d = 0 → a = −2d. Use a + 4d = −4: (−2d) + 4d = −4 ⇒ 2d = −4 ⇒ d = −2. Then a = −2d = 4. Now x100 = a + 99d = 4 + 99(−2) = 4 − 198 = −194. Answer: −194
Explanation:
Given: x5 = a + 4d = −4 → a + 4d = −4. 2x6 + 2x9 = x11 + x13. Compute terms: x6 = a + 5d x9 = a + 8d x11 = a + 10d x13 = a + 12d Plug into (2): 2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d) → 2a + 10d + 2a + 16d = 2a + 22d → 4a + 26d = 2a + 22d → 2a + 4d = 0 → a + 2d = 0 → a = −2d. Use a + 4d = −4: (−2d) + 4d = −4 ⇒ 2d = −4 ⇒ d = −2. Then a = −2d = 4. Now x100 = a + 99d = 4 + 99(−2) = 4 − 198 = −194. Answer: −194
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