Question

Consider a sequence $a_1,a_2,a_3,\dots \dots$ 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 $\left(365-\sum _{n=1}^{10}{a}_n\right)$

Answer 1

Answer: 5

$S=a_1+a_2+a_3+\dots \dots a_n$
$S=a_1+a_2+\dots \dots +a_{n-1}+a_n$
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$a_n=a_1+\left(T_1+T_2+\dots \dots T_{n-1}\right)=3+\frac{n-1}{2}(4+(n-2)2)$
$\Rightarrow a_n=3+(n-1)n=n^2-n+3$
$\therefore \sum _{n=1}^{10}{a}_n=\sum _{n=1}^{10}\left(n^2-n+3\right)=360$
$\therefore 365-\sum _{n=1}^{10}{a}_n=5$