Question

Match the statements in List 1 with those in List 2

	List 1		List 2
A.	If $a,b,c,d$ are in A.P., then (where $a,b,c\&d$ $\in R^{+}$)	1.	$a+d>b+c$
B.	If $a,b,c,d$ are in G.P, then (where $a,b,c\&d$ $\in R^{+}$)	2.	$bc>ad$
C.	The sum of $n$ terms of an A.P. is given by $S_n=bn+c{n}^2$ where $b$ and $c$ are independent of $n$, then	3.	$T_n=b+(2n-1)c$
		4.	Common difference $\left(D\right)=(2b-4)$

• A. $A-2,B-1,C-3$
• B. $A-4,B-2,C-1$
• C. $A-4,B-1,C-3$
• D. $A-4,B-3,C-2$

Answer 1

The correct option is A $A-2,B-1,C-3$

A) $a,b,c,d\to A.P$
$a=a,b=a+D,c=a+2D,d=a+3D\therefore bc=(a+D)(a+2D)=a^2+3aD+{2D}^{2}ad=a(a+3D)=a^2+3aD\Rightarrow bc>ad$
B) $a,b,c.d\to G.P$
$a=a,b=ar,c=a{r}^2,d={ar}^{3}a+d=a+{ar}^3=a(1+r^3)b+c=ar+{ar}^2=ar(1+r)r+r^2<1+r^3\therefore a+d>b+c$
C) $S_n=bn+{cn}^2$
$S_1=T_1=a=b+c{S}_2=T_2+T_1=2b+4c\therefore T_2=2b+4c-b-c\Rightarrow a+d=b+3c\Rightarrow d=b+3c-b-c=+2c\therefore T_n=b+c+(n-1)2c=b+c+2cn-2c=b+c(2n-1)$