Match the following Column–IColumn–IIAThe arithmetic mean of two positive numbers is 6 and their geometric mean G and harmonic mean H satisfy the relation G2+3H=48, then product of the two number isP24077BThe sum of the series 51242+114272+1772.102+... isQ32CIf the first two terms of a Harmonic Progression be 12 and 13, then theHarmonic Mean of the first four terms isR13DGeometric mean of – 4 and – 9S6 T–6

(A) a + b = 12 ab+6aba+b=48 ab+ab2=48∴ ab = 32 (B) S=512+1142.72+1172.102+......... ⇒3S=3.512.42+3.1142.72+3.1772.102+.......... ⇒3S=(4−1).(4+1)12.42+(7−4)(7+4)42.72+(10−7)(10+7)72.102+............⇒3S=42−1212.42+72−4242.72+102−7272.102+............ ⇒3S=1−142+142−172+172−1102+............ ⇒3S=1S=13(C) H.M of 12,13,14,15 is 412+13+14+15=24077(D) Since G.M. lies between the numbers GM=−(−4)×(−9)=−6

Match the following Column–IColumn–IIAThe arithmetic mean of two positive numbers is 6 and their geometric mean G and harmonic mean H satisfy the relation G2+3H=48, then product of the two number isP24077BThe sum of the series 51242+114272+1772.102+... isQ32CIf the first two terms of a Harmonic Progression be 12 and 13, then theHarmonic Mean of the first four terms isR13DGeometric mean of – 4 and – 9S6 T–6

(A) a + b = 12 ab+6aba+b=48 ab+ab2=48∴ ab = 32 (B) S=512+1142.72+1172.102+......... ⇒3S=3.512.42+3.1142.72+3.1772.102+.......... ⇒3S=(4−1).(4+1)12.42+(7−4)(7+4)42.72+(10−7)(10+7)72.102+............⇒3S=42−1212.42+72−4242.72+102−7272.102+............ ⇒3S=1−142+142−172+172−1102+............ ⇒3S=1S=13(C) H.M of 12,13,14,15 is 412+13+14+15=24077(D) Since G.M. lies between the numbers GM=−(−4)×(−9)=−6

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Match the following
Column–I
Column–II
A
The arithmetic mean of two positive numbers is 6 and their geometric mean G and harmonic mean H satisfy the relation $G^{2} + 3 H = 48$ , then product of the two number is
P
$\frac{240}{77}$
B
The sum of the series $\frac{5}{1^{2} 4^{2}} + \frac{11}{4^{2} 7^{2}} + \frac{17}{7^{2} {.10}^{2}} + ...$ is
Q
32
C
If the first two terms of a Harmonic Progression be $\frac{1}{2}$ and $\frac{1}{3}$ , then the
Harmonic Mean of the first four terms is
R
$\frac{1}{3}$
D
Geometric mean of – 4 and – 9
S
6

T
–6

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$A \to r; B \to q; C \to p; D \to t$

$A \to q; B \to r; C \to t; D \to p$

$A \to p; B \to q; C \to r; D \to s$

$A \to q; B \to r; C \to p; D \to t$

answer is C.

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Detailed Solution

(A) a + b = 12
   $a b + \frac{6 a b}{a + b} = 48$
   $a b + \frac{a b}{2} = 48 ∴$ ab = 32
(B)    $S = \frac{5}{1^{2}} + \frac{11}{4^{2} {.7}^{2}} + \frac{11}{7^{2} {.10}^{2}} + .........$
$\Rightarrow 3 S = \frac{3.5}{1^{2} {.4}^{2}} + \frac{3.11}{4^{2} {.7}^{2}} + \frac{3.17}{7^{2} {.10}^{2}} + ..........$
$\Rightarrow 3 S = \frac{(4 - 1) . (4 + 1)}{1^{2} {.4}^{2}} + \frac{(7 - 4) (7 + 4)}{4^{2} {.7}^{2}} + \frac{(10 - 7) (10 + 7)}{7^{2} {.10}^{2}} + ............$ $\Rightarrow 3 S = \frac{4^{2} - 1^{2}}{1^{2} {.4}^{2}} + \frac{7^{2} - 4^{2}}{4^{2} {.7}^{2}} + \frac{10^{2} - 7^{2}}{7^{2} {.10}^{2}} + ............$

$\Rightarrow 3 S = 1 - \frac{1}{4^{2}} + \frac{1}{4^{2}} - \frac{1}{7^{2}} + \frac{1}{7^{2}} - \frac{1}{10^{2}} + ............$
$\Rightarrow 3 S = 1 S = \frac{1}{3}$
(C) H.M of $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$ is $\frac{4}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}} = \frac{240}{77}$
(D) Since G.M. lies between the numbers $G M = - \sqrt{(- 4) \times (- 9)} = - 6$

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