Question

Match the statements in List 1 with those in List 2

	List 1		List 2
A.	Sum of $n$ A.M.'s between a and b is	1.	$(\sqrt{ab}{)}^n$
B.	Product of $n$ G.M.'s between a and b is	2.	$2Apq$
C.	If A, G, H are A.M., G.M., H.M, between the same two numbers, such that $A-G=15$ and $A-H=27$, then the numbers are	3.	$\frac{G_1{G}_2}{H_1{H}_2}$
D.	$A_1,A_2,G_1,G_2$ and $H_1,H_2$ are respective two A.M.'s two G.M's and two H.M.'s between the same two numbers, then $\frac{A_1+A_2}{H_1+H_2}=$	4.	$\frac{n(a+b)}{2}$
		5.	120, 30

• A. $A-4,B-5,C-2,D-3$
• B. $A-3,B-1,C-2,D-5$
• C. $A-4,B-1,C-5,D-3$
• D. $A-3,B-4,C-2,D-1$

Answer 1

The correct option is C $A-4,B-1,C-5,D-3$

(a) Sum of $n$ A.Ms. $=n\times$single A.M.
$A_1+A_2......A_n=n\left(\frac{a+b}{2}\right)$ formulae.

(b)Product of $n$ G.M. $=(singleG.M.{)}^n$
for example $G_1{G}_2=(\sqrt{ab}{)}^2=ab$ formulae
(c) Let the number be a, b and their A.M., G.M., and H.M. be denoted by A, G, and H respectively. Also we know that A, G, H are in G.P.
or $G^2=AH$ .....(1)
Since $A-G=15$ and $A-H=27$
$(A-15{)}^2=G^2=AH$ by (1) $=A(A-27)$
or $-30A+225=-27A$or $3A=225$
$\therefore A=75=\frac{a+b}{2}$ $\therefore a+b=150$ ......(2)
Since $A-G=15$ $\therefore 75-G=15$
or $G=60=\sqrt{ab}$
$ab=3600$ .......(3)
hence from (2) and (3) we conclude that $a$ and $b$ are the roots of $t^2-150t+3600=0$
or $(t-120)(t-30)=0$ $\therefore t=120,30$
Hence the two number are $120$ and $30$.
(d)$\frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},\frac{1}{b}$are in A.P.
$\therefore \frac{1}{H_1}+\frac{1}{H_2}=\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$
or $\frac{H_1+H_2}{H_1{H}_2}=\frac{A_1+A_2}{G_1{G}_2}$etc.