test","extracted_subject":"MATH","extracted_chapter":"PROGRESSIONS","extracted_topic":"ARITHMETIC PROGRESSION","questionPreviewUrl":"https://cdn.pureessence.tech/question-previews/packed_canvases/canvas_1943.png?top_left_x=1064&top_left_y=1605&width=532&height=262&app=solveeit"},"seo_metadata":{"keywords":[]},"doubt_solver_data":{"concept_summary":[]},"embeddings":[],"is_hidden":false,"_id":"66d23e72425d76b7ab31ebcb","slug":"let-xsub1sub-xsub2sub-xsub3subbe-positive-integers","content":"Let x₁, x₂, x₃…..be positive integers in AP such that\n\nx₁ + x₂ + x₃ = 12 & x₄ + x₅ = 14 then x₇ =","options":[{"_id":"68aa85117f15af4716d9d153","text":"10","sequence":0,"isCorrect":true},{"_id":"68aa85117f15af4716d9d154","text":"7","sequence":1,"isCorrect":false},{"_id":"68aa85117f15af4716d9d155","text":"13","sequence":2,"isCorrect":false},{"_id":"68aa85117f15af4716d9d156","text":"None","sequence":3,"isCorrect":false}],"correct_answer":"10","explanation":"a + a + d + a + 2d = 12 ̃

d = 6/5\n\na + 3d + a + 4d = 14 ̃ 2a + 7d = 14 a = 14/5\n\n\\\\ x₇ = a + 6d = \\(\\frac { 14 } { 5 }\\) + \\(\\frac { 36 } { 5 }\\) = 10","question_type":"single_choice","appeared_in":[],"created_at":"2024-08-30T21:49:38.072Z","__v":0,"vector_store_id":"32c42f54-a782-409b-a252-2a8f2a3e51e6","updated_at":"2024-08-30T21:49:38.072Z","isStarred":false},"params":{"questionSlug":"let-xsub1sub-xsub2sub-xsub3subbe-positive-integers"}}]}]]

Question: Let x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>…..be positive integers in AP such that x<sub>1</sub...

Solution