Question

One or more than one correct:

1. If for an increasing GP, a,b,c and d are $l^{th}$, $m^{th}$, $n^{th}$ and $p^{th}$ terms respectively and roots of equation $(a^2+b^2+c^2)x^2-2(ab+bc+cd)x + b^2+c^2+d^2 = 0$ are real, then-
2. Least value of $\frac{x^2y^2-2x^2y+2x^2+2xy-2x+1}{x^2y+x}$ is $\lambda$, then {where $x,y \in R^+$, $x^2y + x \neq 0$}
3. Let $\frac{a}{b}, ab, a - b$ and a + b are first 4 terms of an A.P. in order $(a,b \in R_0)$, then-
4. Let $(a_n), n = 1,2,3....$ be the sequence of real numbers such that $a_1 = 2$ and $a_n = (\frac{n+1}{n-1})(a_1 + a_2 + ... + a_{n-1}), (n \geq 2)$, then- [where [.] denotes greatest integer function]
5. The sum of infinite series $\frac{1}{3} + \frac{3}{3.7} + \frac{5}{3.7.11} + \frac{7}{3.7.11.15} + ....$ is equal to
6. If a, b, c be three unequal positive quantities in H.P., then -
7. If $2a^2, b^2, 2c^2$ are three distinct numbers in harmonic progression then-

• A. $l,m,n,p$ are in G.P
• A. $\lambda \in (0,1)$
• A. common difference of A.P is $-\frac{6}{5}$
• A. $[\frac{a_{2016}}{2^{2016}}] = 1008$
• A. $\frac{1}{2}$
• A. $a^{100} + c^{100} > 2b^{100}$
• A. $4a^2-b^2, b^2, 4c^2-b^2$ are in G.P.
• B. $l,m,n,p$ are in A.P
• B. $\lambda \in [1,3)$
• B. common difference of A.P is $-\frac{5}{6}$
• B. $[\frac{a_{2016}}{2^{2016}}] = 2016$
• B. $\frac{1}{6}$
• B. $a^3 + c^3 > 2b^3$
• B. $4a^2-b^2, b^2, 4c^2-b^2$ are in A.P.
• C. $b^2+c^2 = ac + bd$
• C. $\lambda \in [3,4]$
• C. fifth term of A.P. will be $-\frac{117}{40}$
• C. $a_{2015}$ is divisible by $2^{2019}$
• C. $\frac{4}{3}$
• C. $a^5 + b^5 < 2b^5$
• C. $\frac{ab}{c}, \frac{ac}{b}, \frac{bc}{a}$ are in A.P.
• D. $b^2+c^2 = ad + bc$
• D. $\lambda \in (4,7)$
• D. $ab = \frac{27}{40}$
• D. If $a_1 + a_2 + ....... + a_n > 1008$, then least value of n is 8.
• D. $\frac{8}{3}$
• D. $a^2 + c^2 > 2b^2$
• D. $\frac{ab}{c}, \frac{2ac}{b}, \frac{bc}{a}$ are in A.P.

Answer 1

Answer: (A), (A), (A), (A), (A), (A), (B), (B), (C), (C), (C), (D), (D), (D), (D)

1. (B), (C); 2. (A); 3. (A), (C), (D); 4. (A), (C), (D); 5. (A); 6. (A), (B), (D); 7. (A), (D)

Answer 2

Explanation for each part:

1. The condition for real roots implies the equality in Cauchy-Schwarz inequality, which means $a,b,c,d$ are in GP. This leads to $l,m,n,p$ being in A.P. ($l-m=m-n=n-p$). If $a,b,c,d$ are in GP, then $b=ar, c=ar^2, d=ar^3$. Substituting these into $b^2+c^2=ac+bd$ gives $a^2r^2+a^2r^4 = a(ar^2)+ar(ar^3) = a^2r^2+a^2r^4$, which is true.

2. The expression simplifies to $y+\frac{1}{x}-2+\frac{2x}{xy+1}$. Let $A=y+\frac{1}{x}$. The expression becomes $A-2+\frac{2}{A}$. By AM-GM, $A+\frac{2}{A} \ge 2\sqrt{2}$. Thus, the minimum value is $2\sqrt{2}-2 \approx 0.828$, which is in $(0,1)$.

3. Let the common difference be $d'$. $d' = (a+b)-(a-b) = 2b$. Also, $d' = (a-b)-ab$ and $d' = ab - a/b$. Solving these equations simultaneously yields $b = -3/5$ (since $b \ne 1$). Then $d' = 2(-3/5) = -6/5$. $a = 2b^2/(b^2-1) = -9/8$. $ab = (-9/8)(-3/5) = 27/40$. The fifth term is $T_5 = (a+b)+d' = (-9/8-3/5) + (-6/5) = -69/40 - 48/40 = -117/40$.

4. From $a_n = (\frac{n+1}{n-1})S_{n-1}$ and $a_{n-1} = (\frac{n}{n-2})S_{n-2}$, using $S_{n-1} = S_{n-2}+a_{n-1}$, we derive the recurrence $a_n = \frac{2(n+1)}{n}a_{n-1}$ for $n \ge 2$. Solving this recurrence with $a_1=2$ gives $a_n = 2^{n-1}(n+1)$.

   (A) $a_{2016}/2^{2016} = 2^{2015}(2017)/2^{2016} = 2017/2 = 1008.5$. $[\dots]=1008$.

   (C) $a_{2015} = 2^{2014}(2016) = 2^{2014}(32 \cdot 63) = 2^{2014}(2^5 \cdot 63) = 2^{2019} \cdot 63$. This is divisible by $2^{2019}$.

   (D) The sum $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k+1)2^{k-1}$. This is an Arithmetico-Geometric series sum, which evaluates to $n \cdot 2^n$. We need $n \cdot 2^n > 1008$. $7 \cdot 2^7 = 896$, $8 \cdot 2^8 = 2048$. So the least $n$ is 8.

5. The general term is $T_n = \frac{2n-1}{3 \cdot 7 \cdot 11 \cdots (4n-1)}$. Let $D_n = 3 \cdot 7 \cdot 11 \cdots (4n-1)$. We can write $T_n = \frac{1}{2} \left( \frac{1}{D_{n-1}} - \frac{1}{D_n} \right)$ by defining $D_0=1$. This is a telescoping series. The sum is $\frac{1}{2} \left( \frac{1}{D_0} - \lim_{N \to \infty} \frac{1}{D_N} \right) = \frac{1}{2}(1-0) = \frac{1}{2}$.

6. If $a,b,c$ are in H.P., then $b = \frac{2ac}{a+c}$. Since $a,b,c$ are unequal positive quantities, $a \ne c$. By Power Mean Inequality, for $n \ge 1$, $\frac{a^n+c^n}{2} > (\frac{a+c}{2})^n$. Also, $\frac{a+c}{2} > b$. Combining these, $\frac{a^n+c^n}{2} > b^n \implies a^n+c^n > 2b^n$. This holds for $n=100, 3, 2$. So (A), (B), (D) are correct. (C) $a^5+b^5 < 2b^5 \implies a^5 < b^5 \implies a<b$, which is not always true.

7. If $2a^2, b^2, 2c^2$ are in H.P., then $\frac{1}{2a^2}, \frac{1}{b^2}, \frac{1}{2c^2}$ are in A.P. This implies $\frac{2}{b^2} = \frac{1}{2a^2} + \frac{1}{2c^2} \implies 4a^2c^2 = b^2(a^2+c^2)$.

   (A) If $4a^2-b^2, b^2, 4c^2-b^2$ are in G.P., then $(b^2)^2 = (4a^2-b^2)(4c^2-b^2)$, which simplifies to $b^2(a^2+c^2) = 4a^2c^2$. This matches the condition. So (A) is correct.

   (C) If $\frac{ab}{c}, \frac{ac}{b}, \frac{bc}{a}$ are in A.P., then $2\frac{ac}{b} = \frac{ab}{c} + \frac{bc}{a} \implies 2a^2c^2 = b^2(a^2+c^2)$. This does not match the condition. So (C) is incorrect.

   (D) If $\frac{ab}{c}, \frac{2ac}{b}, \frac{bc}{a}$ are in A.P., then $2(\frac{2ac}{b}) = \frac{ab}{c} + \frac{bc}{a} \implies 4a^2c^2 = b^2(a^2+c^2)$. This matches the condition. So (D) is correct.