IIT JEE Main Maths -Unit 6 -Arithmetic and Geometric progressions- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
From \( a_1 a_5 = a \cdot a r^4 = a^2 r^4 = 28 \) …(1) From \( a_2 + a_4 = ar + ar^3 = ar(1 + r^2) = 29 \) …(2) Dividing (2)\(^2\) by (1): \( a^2 r^2 (1 + r^2)^2 / (a^2 r^4) = (29)^2 / 28 \) gives \( \frac{(1+r^2)^2}{r^2} = \frac{841}{28} \). Solving, \( r = \sqrt{28} \), \( a = \frac{1}{\sqrt{28}} \). Then \( a_6 = a r^5 = \frac{1}{\sqrt{28}} \cdot (28)^{5/2} = 784 \).
Hence, the correct answer is (3) 784.
▶️ Answer/Explanation
\(T_n = S_n – S_{n-1} \Rightarrow T_n = \dfrac{1}{8(2n-1)(2n+1)(2n+3)}\)
\(\sum_{r=1}^n \dfrac{1}{T_r} = \sum_{r=1}^n 8 \left[ \dfrac{1}{(2r-1)(2r+1)} – \dfrac{1}{(2r+1)(2r+3)} \right]\)
This telescopes to \(\dfrac{2}{3}\) as \(n \to \infty\).
The correct answer is \((3)\ \dfrac{2}{3}\).
▶ Answer/Explanation
Ans. (1)
Sol. $a_1, a_2, a_3,\ldots, a_{2k}$ $\rightarrow$ A.P.
$\sum_{r=1}^{k} a_{2r-1} = 40$,
$\sum_{r=1}^{k} a_{2r} = 55$, $a_{2k} – a_1 = 27$
$\frac{k}{2} [2a_1 + (k – 1)2d] = 40$, $\frac{k}{2} [2a_2 + (k – 1)2d] = 55$,
$d = \frac{27}{2k-1}$
$a_1 = \frac{40}{k} – (k – 1)d = \frac{55}{k} – kd$
$d = \frac{15}{k}$ $\Rightarrow \frac{27}{2k-1} = \frac{15}{k}$ $\Rightarrow 9k = 10k – 5$
$\therefore k = 5$
Sol. $a_1, a_2, a_3,\ldots, a_{2k}$ $\rightarrow$ A.P.
$\sum_{r=1}^{k} a_{2r-1} = 40$,
$\sum_{r=1}^{k} a_{2r} = 55$, $a_{2k} – a_1 = 27$
$\frac{k}{2} [2a_1 + (k – 1)2d] = 40$, $\frac{k}{2} [2a_2 + (k – 1)2d] = 55$,
$d = \frac{27}{2k-1}$
$a_1 = \frac{40}{k} – (k – 1)d = \frac{55}{k} – kd$
$d = \frac{15}{k}$ $\Rightarrow \frac{27}{2k-1} = \frac{15}{k}$ $\Rightarrow 9k = 10k – 5$
$\therefore k = 5$
▶️ Answer/Explanation
Ans. (2)
Sol. \(a = 3\)
\(S_4 = \frac{1}{5} (S_8 – S_4)\)
\(\Rightarrow 5S_4 = S_8 – S_4\)
\(\Rightarrow 6S_4 = S_8\)
\(\Rightarrow \frac{4}{2} [2 \cdot 3 + (4-1)d] = \frac{8}{2} [2 \cdot 3 + (8-1)d]\)
\(\Rightarrow 12(6 + 3d) = 4(6 + 7d)\)
\(\Rightarrow 18 + 9d = 6 + 7d\)
\(\Rightarrow d = -6\)
\(S_{20} = \frac{20}{2} [2 \cdot 3 + (20-1)(-6)]\)
\(= 10 [6 – 114]\)
\(= -1080\)