Update

Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है

Subjects
0 reads0 ratings0 helpful

The sequence (3, 8, 15, 24, …) has general term a_n = n² + 2n. Which term is 168?

Advertisement

Answer and explanation

Correct answer: 12th

The governing concept is identifying a term number from an explicit, or general, rule. We need an integer n for which a_n = n² + 2n equals 168. Substitute the possible position n = 12: a_12 = 12² + 2(12) = 144 + 24 = 168. Therefore, 168 is the 12th term, so option C is correct. Checking nearby positions confirms the choice: a_11 = 121 + 22 = 143 and a_13 = 169 + 26 = 195, so neither adjacent position gives 168. The other options result from using an incorrect position or arithmetic.

Related tags

SequencesNth-TermExplicit-RuleTerm-PositionNth TermSequences And ProgressionsMathematicsClass 9 Mcq

Frequently asked questions

What is the correct answer to this question?

12th

Why is this the correct answer?

The governing concept is identifying a term number from an explicit, or general, rule. We need an integer n for which a_n = n² + 2n equals 168. Substitute the possible position n = 12: a_12 = 12² + 2(12) = 144 + 24 = 168. Therefore, 168 is the 12th term, so option C is correct. Checking nearby positions confirms the choice: a_11 = 121 + 22 = 143 and a_13 = 169 + 26 = 195, so neither adjacent position gives 168. The other options result from using an incorrect position or arithmetic.

Which subject and chapter does this question cover?

This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.

Was this question useful?

No ratings yetWrite a review / Rate this question

Student Reviews

No published reviews yet.

Advertisement