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STD 11 - 8. sequence and series — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 8. sequence and series1 Mark
MCQ
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
    $A,B$
  • ✓
    $B,C$
  • C
    $A,C$
  • D
    $A,D$
✓

Answer

Correct option: B.
$B,C$
b
$a _1=7, d =8$

$T _{ n +1}- T _{ n }= a _{ n } \forall n \geq 1$

$S _{ n }= T _1+ T _2+ T _3+\ldots+ T _{ n -1}+ T _{ n }$

$S _{ n }=\quad T _1+ T _2+ T _3+\ldots .+ T _{ n -1}+ T _{ n }$

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 $

$\sum_{k=1}^{20} T_k=10510$

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