Question

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. $\displaystyle\sum_{ k =1}^{20} T_{ k }=10510$
• C. $T_{30}=3454$
• D. $\displaystyle\sum_{ k =1}^{30} T_{ k }=35610$

Answer 1

Answer: (B), (C)

$ 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 }= 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$
$T _{30}=3454$
$ \displaystyle\sum_{ k =1}^{30} T _{ k }=35615$
$ \displaystyle\sum_{ k =1}^{20} T _{ k }=10510$