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 = 4n² + 3n, find the sum from the 61st term to the 90th term.

Answer and explanation

Correct answer: 18090

When S_n represents the sum of the first n terms, the sum from the rth term to the sth term is S_s − S_{r−1}. The required range is 61 through 90, so calculate S_90 − S_60. First, S_90 = 4(90²) + 3(90) = 4(8100) + 270 = 32,670. Next, S_60 = 4(60²) + 3(60) = 4(3600) + 180 = 14,580. Therefore the required sum is 32,670 − 14,580 = 18,090. Option A is correct. Subtracting S_61 instead of S_60 would remove the 61st term, and using the wrong square or coefficient would produce a distractor. The use of r − 1 is the key boundary idea.

Related tags

Arithmetic ProgressionPartial SumRange SumConsecutive TermsFinding The Sum Of The First $N$ Terms Of An ApFinding The Sum Of The First N Terms Of An ApArithmetic Progressions (Ap)Arithmetic Progressions Ap

Was this question useful?

No ratings yetWrite a review / Rate this question

Student Reviews

No published reviews yet.

Add Muft Shiksha to your Home Screen

In Safari, tap Share, then Add to Home Screen.