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Let $a_1, a_2, a_3, \ldots$ be an arithmetic progression with $a_1=7$ and common difference $8$ . Let $T_1, T_2, T_3, \ldots$ be such that $T_1=3$ and $T_{n+1}-T_n=a_n$ for $n \geq 1$. Then, which of the following is/are $TRUE$ ?
$(A)$ $T_{20}=1604$
$(B)$ $\sum_{ k =1}^{20} T_{ k }=10510$
$(C)$ $T_{30}=3454$
$(D)$ $\sum_{ k =1}^{30} T_{ k }=35610$
$A,B$
$B,C$
$A,C$
$A,D$
Solution
$a _1=7, d =8$
$T _{ n +1}- T _{ n }= a _{ n } \forall n \geq 1$
$S _{ n }= T _1+ T _2+ T _3+\ldots+ T _{ n -1}+ T _{ n }$
$S _{ n }=\quad T _1+ T _2+ T _3+\ldots .+ T _{ n -1}+ T _{ n }$
on subtraction
$T_n=T_1+a_1+a_2+\ldots .+a_{n-1}$
$T_n=3+(n-1)(4 n-1)$
$T_n=4 n^2-5 n+4$
$\sum_{k=1}^n T_k=4 \sum n^2-5 \sum n+4 n$
$T_{20}=1504$
$T_{30}=3454$
$\sum_{k=1}^{30} T_k=35615 $
$\sum_{k=1}^{20} T_k=10510$
