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JEE Mains · Maths · STD 11 - 8. sequence and series

જો શ્રેણી \(20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots .\) upto \(n ^{ th }\) પદ સુધીનો સરવાળો \(488\) અને \(n^{\text {th }}\) પદ ઋણ હોય તો 

  1. A \(n ^{\text {th }}\) મુ પદ \(-4 \frac{2}{5}\) થાય 
  2. B \(n =41\)
  3. C \(n^{\text {th }}\) મુ પદ \(-4\) થાય 
  4. D \(n =60\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(n^{\text {th }}\) મુ પદ \(-4\) થાય 

Step-by-step Solution

Detailed explanation

\(S =\frac{100}{5}+\frac{98}{5}+\frac{96}{5}+\frac{94}{5}+\ldots . . n\) \(S _{ n }=\frac{ n }{2}\left(2 \times \frac{100}{5}+( n -1)\left(-\frac{2}{5}\right)\right)=188\) \(n (100- n +1)=488 \times 5\) \(n ^{2}-101 n +488 \times 5=0\) \(n =61,40\)…
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