Question

Consider a sequence $a_1, a_2, a_3, \ldots \ldots$ of real numbers and let the sequence $T_1, T_2, T_3$ ... defined by $T _{ r }= a _{ r +1}- a _{ r }$ be such that its terms are in A.P. with common difference 2 . If $T _1=2$ and $a_1=3$, then find the value of $\left(365-\displaystyle\sum_{n=1}^{10} a_n\right)$

Answer 1

$S = a _1+ a _2+ a _3+\ldots \ldots a _{ n }$
$S = a _1+ a _2+\ldots \ldots+ a _{ n -1}+ a _{ n } $
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$a _{ n }= a _1+\left( T _1+ T _2+\ldots \ldots T _{ n -1}\right)=3+\frac{ n -1}{2}(4+( n -2) 2)$
$\Rightarrow a_n=3+(n-1) n=n^2-n+3 $
$\therefore \displaystyle\sum_{n=1}^{10} a_n=\displaystyle\sum_{n=1}^{10}\left(n^2-n+3\right)=360$
$\therefore 365-\displaystyle\sum_{n=1}^{10} a_n=5$