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

Answer 1

Answer: (B)

$\displaystyle\sum_{ k =1}^{20} T_{ k }=10510$

Answer 2

According to the question,
$\sum\limits_{n=1}^n(T_{n+1}-T_n)=\sum a_n$
$⇒T_{n+1}-T_1=\sum a_n$
$=\frac{n}{2}[2\times7+(n-1)8]$
So, by solving this, we get :
Tn+1 = n(4n + 3) + T1
Tn+1 = 4n2 + 3n + 3
Now, by using this equation the above equation in the options, we get :
In (A),
T20 = 4 × 192 + 3 × 9 + 3
= 1444 + 27 + 3 = 1474
In (B),
$\sum\limits^{19}_{k=0}T_{n+1}=\sum\limits_{k=0}^{19}k(4k+3)+3$
$=\sum\limits_{k=0}^{19}(4k^2+3k+3)=10510$
In (C),
$\sum\limits^{30}_{k=1}T_k=\sum\limits^{29}_{n=0}T_{n+1}=\sum\limits_{n=0}^{29}n(4n+3)+3$
$=4\times(\frac{29\times30\times59}{6})+\frac{3(29\times30)}{2}+90$
$=35615$
In (D),
T30 = 29(4 × 29 + 3) + 3 = 3454
So, we can see that only (B) and (D) are true statements.
Therefore, the correct option is (B) :$\displaystyle\sum_{ k =1}^{20} T_{ k }=10510$ and (D) : $T_{30}=3454$