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

मान लीजिए किसी अनुक्रम का पहला पद \(T_1=6\) है और इसका \(\mathrm{r}^{\text {th }}\) पद \(T_r=3 T_{r-1}+6^r, r=2,3, \ldots . ., n\). यदि इस अनुक्रम के पहले \(\mathrm{n}\) पदों का योग \(\frac{1}{5}\left(n^2-12 n+39\right)\) \(\left(4.6^n-5.3^n+1\right)\) है, तो \(\mathrm{n}\) = ...........

  1. A \(10\)
  2. B \(5\)
  3. C \(6\)
  4. D \(11\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(6\)

Step-by-step Solution

Detailed explanation

\( \mathrm{T}_{\mathrm{r}}=3 \mathrm{~T}_{\mathrm{r}-1}+6^{\mathrm{r}}, \mathrm{r}=2,3,4, \ldots \mathrm{n} \) \( \mathrm{T}_2=3 \cdot \mathrm{T}_1+6^2 \) \( \mathrm{~T}_2=3 \cdot 6+6^2 \) ................(\(1\)) \( \mathrm{~T}_3=3 \mathrm{~T}_2+6^3 \)…
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