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

If \(S_n=300\), what will be the value of \(n\)?

Advertisement

Answer and explanation

Correct answer: 24

For the sum of the first \(n\) natural numbers, \(S_n=n(n+1)/2\). We need \(n(n+1)/2=300\), so \(n(n+1)=600\). Among the options, \(n=24\) gives \(24\times25=600\), and therefore \(S_{24}=24\times25/2=300\). Hence option B is correct. The equation can also be solved as \(n^2+n-600=0\), which factors as \((n+25)(n-24)=0\); the positive natural-number solution is 24. The other options produce sums different from 300, so they do not satisfy the defining formula.

Related tags

Sum Of Natural NumbersQuadratic ReasoningFind NSum Of First N Natural NumbersSequences And ProgressionsMathematicsClass 9 Mcq

Frequently asked questions

What is the correct answer to this question?

24

Why is this the correct answer?

For the sum of the first \(n\) natural numbers, \(S_n=n(n+1)/2\). We need \(n(n+1)/2=300\), so \(n(n+1)=600\). Among the options, \(n=24\) gives \(24\times25=600\), and therefore \(S_{24}=24\times25/2=300\). Hence option B is correct. The equation can also be solved as \(n^2+n-600=0\), which factors as \((n+25)(n-24)=0\); the positive natural-number solution is 24. The other options produce sums different from 300, so they do not satisfy the defining formula.

Which subject and chapter does this question cover?

This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Sum of first n natural numbers.

Was this question useful?

No ratings yetWrite a review / Rate this question

Student Reviews

No published reviews yet.

Advertisement