Let 𝑎₁, 𝑎₂, 𝑎₃,… be an arithmetic progression with $a_{1} = 7$ and common difference 8. Let 𝑇₁, 𝑇₂, 𝑇_3,… 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 ?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

𝑇₃₀ = 3454

$\sum_{k = 1}^{30} T_{k} = 35610$

𝑇₂₀= 1604

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Here $a_{n} = 7 + (n - 1) 8 and T_{1} = 3$
$\begin{array}{ll} Also T_{n + 1} = T_{n} + a_{n} \\ T_{n} = T_{n - 1} + a_{n - 1} \\ ⋮ \\ T_{2} = T_{1} + a_{1} \\ ∴ T_{n + 1} = (T_{n - 1} + a_{n - 1}) + a_{n} \\ = T_{n - 2} + a_{n - 2} + a_{n - 1} + a_{n} \\ ⋮ \\ \Rightarrow T_{n + 1} = T_{1} + a_{1} + a_{2} + \dots + a_{n} \\ \Rightarrow T_{n + 1} = T_{1} + \frac{n}{2} [2 (7) + (n - 1) 8] \\ \Rightarrow T_{n + 1} = T_{1} + n (4 n + 3) \\ ∴ For n = 19 T_{20} = 3 + (19) (79) = 1504 \\ For n = 29 T_{30} = 3 + (29) (119) = 3454 \to (C) \\ 20 \\ \sum_{k = 1}^{2} T_{k} = 3 + \sum_{k = 2}^{20} T_{k} = 3 + \sum_{k = 1}^{19} (3 + 4 n^{2} + 3 n) \\ = 3 + 3 (19) + \frac{3 (19) (20)}{2} + \frac{4 (19) (20) (39)}{6} \\ = 3 + 10507 = 10510 \to (B) \end{array}$
And Similarly $\sum_{k = 1}^{30} T_{k} = 3 + \sum_{k = 1}^{29} (4 n^{2} + 3 n + 3) = 35615$