Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a₄ = a r³ = x … (1)
a₁₀ = a r⁹ = y … (2)
a₁₆ = a r¹⁵ = z … (3)
Dividing (2) by (1), we obtain
\(\frac{y}{x}=\frac{a r^{9}}{a r^{3}} \Rightarrow \frac{y}{x}=r^{6}\)
Dividing (3) by (2), we obtain
\(\frac{z}{y}=\frac{a r^{15}}{a r^{9}} \Rightarrow \frac{z}{y}=r^{6}\)
\(\therefore \frac{y}{x}=\frac{z}{y}\)
Thus x,y,z are in G.P.

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 A be the first term and D be the common difference of the given AP. Then,
\({ T }_{ p }=a\Rightarrow A+(p-1)D=a\quad \quad \quad \quad \quad ...(i)\)
\(and\quad { T }_{ q }=b\Rightarrow A+(q-1)D=b\quad ...(ii)\)
On subtracting Eq.(ii) from Eq.(i), we get
\((p-q)D=a-b\Rightarrow D=\frac { a-b }{ p-q } \quad \quad \quad ...(ii)\)
On adding Eqs.(i) and (ii), we get
\(2A+(p+q-2)D=(a+b)\)
\( \Rightarrow 2A+pD+qD-2D=a+b\)
\(\Rightarrow A+pD+qD-D=a+b+D\)
\(\Rightarrow 2A+(p+q-1)D=a+b+D\)
\(\Rightarrow (p+q-1)D=a+b+\left( \frac { a-b }{ p-q } \right) \) [from Eq.(iii)]  ....(iv)
Now,  \({ S }_{ p+q }=\frac { p+q }{ 2 } [2A+(p-q-1)D]\)
\(\frac { p+q }{ 2 } \left[ a+b+\frac { a-b }{ p-q } \right] \) [from Eq.(iv)
Hence proved.

23

Given AP is 3, 7, 11,.....407
Here, first term, a = 3 and common difference, d = 4 and nth term, T_n = 407
Also, T_n = a + (n - 1)d
\(\therefore \)  407  = 3 + (n -1) 4 \(\Longrightarrow \)  407 = 3 + 4n -4
\(\Longrightarrow \) 4n = 407 - 3 + 4  \(\Longrightarrow \)  4n = 408 \(\Longrightarrow \)  n = 102
Hence, there are 102 terms in the given AP.
Now, 18th term from the end = [n - m +1]th term from the beginning = [102 -18 +1]th term from the beginning = 85th term from the beginning = 3 + (85 - 1)4 = 339.

Here, S_n = 3n²+5n and a_m = 164.
Replacing n by (n-1) in S_n , we have:
S_n-1 = 3(n-1)² + 5(n-1)
=3(n²-2n+1)+5n-5
=3n²-6n+3+5n-5=3n²-n-2
We know that a_n = S_n - S_n-1
\(\therefore\) a_n = 3n² + 5n-(3n²-n-2)
= 3n²+5n-3n²+n+2=6n+2
Replacing n by m is a_n, we have
\(\therefore\) a_m = 6m + 2
Now  6m + 2 = 164
\(\therefore\) 6m = 164 - 2
\(\Rightarrow m=\frac { 162 }{ 6 } \Rightarrow m=27\)

Two digit numbers are: 10, 11, 12, ..., 99
\(\therefore\)  Two digit numbers which when divided by 4, yield 1 as remainder are: 13, 17, 21, ..., 97.
Here, a = 13, d = 17 -13 = 4 and an = 97
.\(\because\)  a_n = a + (n - l)d
\(\therefore\)  97 = 13 + (n - 1) x 4
\(\Rightarrow\) n - 1 = \(\frac { 97-13 }{ 4 } \)
\(\Rightarrow\) n - 1= \(\frac { 84 }{ 4 } =21\)
\(\Rightarrow\) n = 21 + 1 = 22
Sn = \(\frac { n }{ 2 } \) (a+a_n)
\(\Rightarrow\) S₂₂ = \(\frac { 22 }{ 2 } \) (13+97)
= 11(110) = 1210

1+1\(\frac{1}{7}\)+1\(\frac{2}{7}\)+.........

We have, a=3, l=57, and n=20
\(\because { S }_{ n }=\frac { n }{ 2 } [a+l]\)
\(\therefore { S }_{ 20 }=\frac { 20 }{ 2 } [3+57]=10\times 60=600\)

Here a₁ = 3, and a_n = 3a_n-1 + 2.
Putting n = 2, 3, 4 and 5, we have
a₂ = 3a_2-1 + 2=3a₁ + 2 = 3 \(\times\) 3 + 2
[\(\because\) a₁ = 3]
= 9 + 2 = 11
a₃ = 3a_3-1+2=3a₂ + 2 = 3 \(\times\) 11 + 2
= 33 + 2 = 35   [\(\because\) a₂ = 11]
a₄ = 3a_4-1+2=3a₃+2=3 \(\times\)35 + 2
[ \(\because\) a₃ = 35]
= 105 + 2 = 107
a₄ = 3a_5-1 + 2 = 3a₄+2 = 3 \(\times\) 107 + 2
[\(\because\) a₄ = 107]
= 321 + 2 = 323
Thus, first five terms of the sequences are 3, 11, 35, 107 and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323+ ....

Here, \({ a }_{ n }=\frac { n }{ n+1 } \)
Putting n = 1, 2, 3, 4 and 5, we have
\({ a }_{ 1 }=\frac { 1 }{ 1+1 } =\frac { 1 }{ 2 } { a }_{ 2 }=\frac { 2 }{ 2+1 } =\frac { 2 }{ 3 } \)
\({ a }_{ 3 }=\frac { 3 }{ 3+1 } =\frac { 3 }{ 4 } { a }_{ 4 }=\frac { 4 }{ 4+1 } =\frac { 4 }{ 5 } \)
\({ a }_{ 5 }=\frac { 5 }{ 5+1 } =\frac { 5 }{ 6 } \)
Thus, the first five terms of sequence are  \(\frac { 1 }{ 2 }\),\(\frac { 2 }{ 3 }\),\(\frac { 3 }{ 4 }\), \(\frac { 4 }{ 5 }\) and \(\frac { 5 }{ 6 }\)

Rs.39100

\(0.\bar { 15 } \) = 0.15555..=0.1+0.05+0.005+0.0005+...\(\infty \)
 \(=0.1+\left[ \frac { 5 }{ 100 } +\frac { 5 }{ 1000 } +\frac { 5 }{ 10000 } +...\infty \right] =0.1+\frac { \left( \frac { 5 }{ 100 } \right) }{ 1-\frac { 1 }{ 10 } } =\frac { 7 }{ 45 } \)

A = \(\frac { a+b }{ 2 } \) and G₁ c ar, G₂ = ar²
Where, \(r=\left( \frac { b }{ a } \right) ^{ \frac { 1 }{ 2+1 } }=\left( \frac { b }{ a } \right) ^{ 1/3 }\) ....(i)
\(\therefore \)  \(\frac { G\overset { 2 }{ 1 } }{ { G }_{ 2 } } +\frac { G\overset { 2 }{ 2 } }{ { G }_{ 1 } } \)   
= \(\frac { { a }^{ 2 }{ r }^{ 2 } }{ { ar }^{ 2 } } +\frac { { a }^{ 2 }{ r }^{ 4 } }{ ar } \)

T_z = 7\(\frac { 3 }{ 2 } \)  \(\Longrightarrow \)  a + d = \(\frac { 31 }{ 4 } \)  and T₃₁ = \(\frac { 1 }{ 2 } \)  \(\Longrightarrow \)  a + 30d = \(\frac { 1 }{ 2 } \)
On solving, we get   d = \(\frac { -1 }{ 4 } \) , a= 8
Now, 8 + (n - 1) \(\left( \frac { -1 }{ 4 } \right) \) = \(\frac { -13 }{ 2 } \) = 8, 59