ar^8-1=192\(\Rightarrow \)a\(\times ({ 2) }^{ 7 }=192\Rightarrow a=\frac { 3 }{ 2 } \)

We know that,  AM > GM
\(\therefore \frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2 } >\sqrt { { a }^{ 2 }{ b }^{ 2 } } \Rightarrow \frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2 } >ab\)  ...(i)
\(similarly, \frac { { b }^{ 2 }+{ c }^{ 2 } }{ 2 } >\sqrt { { b }^{ 2 }{ c }^{ 2 } } \Rightarrow \frac { { b }^{ 2 }+{ c }^{ 2 } }{ 2 } >bc\)  ...(ii)
\(and \ \frac { { c }^{ 2 }+{ a }^{ 2 } }{ 2 } >\sqrt { c^{ 2 }{ a }^{ 2 } } \Rightarrow \frac { { c }^{ 2 }+{ a }^{ 2 } }{ 2 } >ca\)  ...(iii)
On adding Eqs. (i) and (iii), we get
\(\frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2 } +\frac { { b }^{ 2 }+{ c }^{ 2 } }{ 2 } +\frac { { c }^{ 2 }+{ a }^{ 2 } }{ 2 } >ab+bc+ca\)
\(\Rightarrow \)  a² + b² + c² > ab + bc + ca
Hence proved              

Let T_n < 0
\(\Longrightarrow \) \(\left[ 19+(n-1)\left( -\frac { 4 }{ 5 } \right) \right] \)< 0 \(\Longrightarrow \) n > 24 \(\frac { 3 }{ 4 } \) = 25th term

Let the GP a,ar,ar²,ar³,...
According to the given condition,
Sum of first three terms = a+ar+ar²=16  ...(i)
And sum of next three terms =ar³+ar⁴+ar⁵=128 ....(ii)
On dividing Eq.(i) by Eq (ii),we get
\(\frac { { a+ar+ar }^{ 2 } }{ ar^{ 3 }+{ ar }^{ 4 }+{ ar }^{ 5 } } =\frac { 16 }{ 128 } \) 
\(\Rightarrow \frac { a(1+r+{ r }^{ 2 }) }{ { ar }^{ 3 }(1+r+{ r }^{ 2 }) } =\frac { 1 }{ 8 } \)
\( \Rightarrow \left( { \frac { 1 }{ r } } \right) ^{ 3 }=\left( { \frac { 1 }{ 2 } } \right) ^{ 3 }\)
On comparing the basr of the power 3 from both sides,
we get \(\Rightarrow 7a=16\Rightarrow a=\frac { 16 }{ 7 } \) 
On putting r =2 in Eq(i) we get
a+2a+4a=16\(\Rightarrow 7a=16\Rightarrow a=\frac { 16 }{ 7 } \) 
Now,sum of terms ,\({ S }_{ n }=\frac { a\left( { r }^{ n }-1 \right) }{ r-1 } \)
\(=\frac { \frac { 16 }{ 7 } \left( { 2 }^{ n }-1 \right) }{ 2-1 } =\frac { 16 }{ 7 } \left( { 2 }^{ n }-1 \right) \) 
Hence,\(a=\frac { 16 }{ 7 } ,r=2\) and \({ S }_{ n }=\frac { 16 }{ 7 } \left( { 2 }^{ n }-1 \right) \)

In the given AP, let the first term = a and the common difference = d.
According to the question, a_m = \(\frac { 1 }{ n } \)  and a_n = \(\frac { 1 }{ m } \) 
\(\therefore \)  a + (m - 1)d = \({ 1 }/{ n }\) ......(i)
and   a + (n -1)d = \({ 1 }/{ m }\) .....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(m - n)d = \(\left( \frac { 1 }{ n } -\frac { 1 }{ m } \right)\) \(\Longrightarrow \) d = \(\frac { 1 }{ mn } \) 
On putting d = \(\frac { 1 }{ mn } \) in Eq. (i), we get
\(a+\frac { (m-1) }{ mn } \) = \(\frac { 1 }{ n } \)  \(\Longrightarrow \) a = \(\left\{ \frac { 1 }{ n } -\frac { (m-1) }{ mn } \right\} \) = \(\frac { 1 }{ mn } \),
Now, mnth term = a + (mn - 1)d
= \(\left\{ \frac { 1 }{ n } -\frac { (mn-1) }{ mn } \right\} \) = \(\frac { mn }{ mn } \) = 1
Hence, the mnth term of the given P is 1

The 16th day of strike , the OPD will have no patient.

Sequence is -\(\frac { 5 }{ 2 }\), \(\frac { 11 }{ 2 }\), \(\frac { 21 }{ 2 }\), \(\frac { 35 }{ 2 }\), -\(\frac { 53 }{ 2 }\),.....
Series is -\(\frac { 5 }{ 2 }\)+\(\frac { 11 }{ 2 }\)+\(\frac { 21 }{ 2 }\)+\(\frac { 35 }{ 2 }\)-\(\frac { 53 }{ 2 }\).......

b,c,d are in GP.
\(\Rightarrow \) C² = bd...(ii)
Similarly, \(\frac { 1 }{ c } ,\frac { 1 }{ d } ,\frac { 1 }{ e } \)are in AP.
\(\Rightarrow \)  \(\frac { 2 }{ d } =\frac { 1 }{ c } +\frac { 1 }{ e } \)
\(\Rightarrow \)  \(d=\frac { 2ce }{ c+e } \)
On putting the values of b and d from Eq.(i) and (iii), in Eq.(ii), we get
\({ c }^{ 2 }=\left( \frac { a+c }{ 2 } \right) \times \left( \frac { 2ce }{ c+e } \right) \Rightarrow { c }^{ 2 }=ae\)
Therefore, a,c,e are in GP.

 Let  A₁,A_2,A₃,A₄,.....,A_m be m AM's between 1 and 31 are in AP.
Therefore, 1,A₁,A_2,A₃,A₄,.....,A_m 31 are in AP.
Here, the total number of terms is m+2 and
T_m+2 =31
\(\Rightarrow\) 1+ ( m + 2 - 1) d =31 \(\Rightarrow\)( m + 1 ) d = 30
\(\Rightarrow\) d = \(\frac{30}{m+1} \)  .....(i)
\(\therefore \)  A₇= T₈ = a + 7d
=1+7\( \times \)\(\frac{30}{m+1}\) =  \(\frac  {m + 211}{ m + 1 }\)   [from Eq.(i)]
A_m-1 = T_m=1+ ( m - 1 ) d = 1+( m - 1 ) d =1+( m - 1 ) \(\frac{30}{m+1}\)
= \( \frac{m+1+30m-30}{m+1} \)   [from Eq.(i)]
= \(\frac{31m-29}{m+1}\)
\(\therefore\)  \(\frac{A7}{Am-1}\) = \( \frac { (m+211) / (m+1)   }{ (31m-29) / m+1  }\)  =  \(\frac{m+211}{31m-29} \)  \(\Rightarrow\) m = 14

Let a and b two numbers and A₁, A₂,.........A_2n be 2n ( even numbers ) AMs.Then a + b = \(\frac{13}{6} \) and 
A₁ + A₂ +......+ A_2n = 2n \((\frac {a+b}{2})\) = \((\frac{13}{6})\)  n
According to the question, 
A₁ + A₂ +......+ A_2n=2n+1 \(\Longrightarrow \)  n=6
Ans. 12

Let 'a' be the 1^st term and 'd' be the common difference of the given AP.
\(\therefore\) S_n = \(\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\therefore\) S_n = \(\frac { 2n }{ 2 } \left[ 2a+\left( 2n-1 \right) d \right] \)
\(\therefore\)  S_n = \(\frac { 3n }{ 2 } \left[ 2a+(3n-1)d \right] \)
Now S₂ - S₁ = \(\frac { 2n }{ 2 } \left[ 2a+(2n-1)d \right] -\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)
= \(\frac { n }{ 2 } \left[ 4a+4nd-2d-2a-nd+d \right] \)
= \(\frac { n }{ 2 } \left[ 2a+3nd-d \right] \)
= \(\frac { n }{ 2 } [2a+(3n-1)d]\)
3(S₂-S₁) = \(\frac { 3n }{ 2 } [2a+(3a-1)d]\) = S₃
[\(\therefore\) S₃ = \([2a+(3a-1)d]\)
Thus, S₃ = 3 (S₂-S₁)

Then  a + b \(6\sqrt { ab } \) \(\Rightarrow\) \(\frac { a+b }{ 2\sqrt { ab } } =\frac { 3 }{ 1 } \)
Applying componendo and dividendo, we get
\(\frac { a+b+2\sqrt { ab } }{ a+b-2\sqrt { ab } } =\frac { 3+1 }{ 3-1 } \)
\(\Rightarrow\) \(\frac { \left[ \sqrt { a } +\sqrt { b } \right] ^{ 2 } }{ \left[ \sqrt { a } -\sqrt { b } \right] ^{ 2 } } =\frac { 4 }{ 2 } \)
\(\Rightarrow\) \(\frac { \left( \sqrt { a } +\sqrt { b } \right) ^{ 2 } }{ \left( \sqrt { a } -\sqrt { b } \right) ^{ 2 } } =\frac { 2 }{ 1 } \)
\(\Rightarrow\) \(\frac { \sqrt { a } +\sqrt { b } }{ \sqrt { a } -\sqrt { b } } =\frac { \sqrt { 2 } }{ 1 } \)
Again applying componendoand dividendowe have:
\(\frac { \sqrt { a } +\sqrt { b } +\sqrt { a } -\sqrt { b } }{ \sqrt { a } +\sqrt { b } -\sqrt { a } -\sqrt { b } } =\frac { \sqrt { 2 } +1 }{ \sqrt { 2 } -1 } \)
\(\Rightarrow\)  \(\frac { \sqrt { a } }{ \sqrt { b } } =\frac { \sqrt { 2 } +1 }{ \sqrt { 2 } -1 } \)
Squaring both sides, we get
\(\frac { a }{ b } =\frac { 2+1+2\sqrt { 2 } }{ 2+1-2\sqrt { 2 } } \) \(\Rightarrow\) \(\frac { a }{ b } =\frac { 3+2\sqrt { 2 } }{ 3-2\sqrt { 2 } } \)
Thus, the numbers are in the ratio
\(\left( 3+2\sqrt { 2 } \right) :\left( 3-2\sqrt { 2 } \right) \)

We know the G.M between 'a' and 'b' is \(\sqrt { ab } \)
\(\therefore\)  \(\frac { { a }^{ n+1 }+b^{ n+1 } }{ { a }^{ n }+{ b }^{ n } } ={ a }^{ \frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } }\)
\(\Rightarrow\) \({ a }^{ n+1 }+{ b }^{ n+1 }={ a }^{ n+\frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } }+{ a }^{ \frac { 1 }{ 2 } }{ b }^{ n+\frac { 1 }{ 2 } }\)
\(\Rightarrow\) \({ a }^{ n+1 }-{ a }^{ n+\frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } }={ a }^{ \frac { 1 }{ 2 } }{ b }^{ n+\frac { 1 }{ 2 } }-{ b }^{ n+1 }\)
\(\Rightarrow\) \({ a }^{ n+\frac { 1 }{ 2 } }\left( { a }^{ \frac { 1 }{ 2 } }-{ b }^{ \frac { 1 }{ 2 } } \right) ={ b }^{ n+\frac { 1 }{ 2 } }\left( { a }^{ \frac { 1 }{ 2 } }-{ b }^{ \frac { 1 }{ 2 } } \right) \)
\(\Rightarrow\) \({ a }^{ n+\frac { 1 }{ 2 } }=b^{ n+\frac { 1 }{ 2 } }\)
\(\Rightarrow\) \(\frac { { a }^{ n+\frac { 1 }{ 2 } } }{ { b }^{ n+\frac { 1 }{ 2 } } } =1\)
\(\Rightarrow\) \(\left( \frac { a }{ b } \right) ^{ n+\frac { 1 }{ 2 } }=\left( \frac { a }{ b } \right) ^{ 0 }\)
\(\Rightarrow\) \(n+\frac { 1 }{ 2 } =0\) \(\Rightarrow\) \(n=-\frac { 1 }{ 2 } \)