Sequences and Series

$\mathrm{}488=\frac{n}{2}\left. [2\left(\frac{100}{5}\right)+(n-1)\left(\frac{2}{5}\right)\right]$  

$488=\frac{n}{2}(101-n)$

$\Rightarrow \mathrm{}{\mathrm{n}}^2-101\mathrm{n}+2440=0$

$\Rightarrow \mathrm{n}=61$ or 40

For $n=40\mathrm{}\Rightarrow \mathrm{}{T}_n>0$  

For $\mathrm{n}=61\mathrm{}\Rightarrow \mathrm{}{\mathrm{T}}_{\mathrm{n}}<0$

${\mathrm{T}}_{\mathrm{n}}=\frac{100}{5}+(61-1)\left(-\frac{2}{5}\right)=-4$

(0.16)^log? (1/3 + 1/3² + . to ∞)
= (4/25)^log? (1/2)
= ( (5/2)? ² )^log? /? (1/2) = (5/2)^ (-2 log? /? (1/2)
= (5/2)^ (log? /? ( (1/2)? ² ) = 4

S = (2 . ¹P? - 3 . ²P? + 4 . ³P? upto 51 terms) + (1! - 2! + 3! - . upto 51 terms)
∴ [? ? P_ (n-1) = n!]
= (2! - 3! + 4! + 52!) + (1! - 2! + 3! - 4! + . . + (51)!)
= 1! + 52!

Sum of 1st 25 terms = sum of its next 15 terms
? (T? + . + T? ) = (T? + . + T? )
? (T? + . + T? ) = 2 (T? + . + T? )
? 40/2 [2*3 + (39d)] = 2 * 25/2 [2*2 + 24 d]
? d = 1/6