Given, depreciation in value of machine = 20%
After each year the value of the machine is 80 %
(100-20)% of its value of the previous year, so at the end of 5 yrs, the machine will depreciate as many times as 5.
Hence, we have to find the 6th terms of the G P whise first term a₁ is 1250 and common ratio r is 8.
Hence, value at the end 5 y6rs.
=t₆=a_rr⁵
=1250(0.8)⁵=409.6

Given a, b, c, d are in GP.
and we know that AM > GM then for the first three terms
 \(\frac { a+c }{ 2 } >b\)   [\(\because \sqrt { ac } =b\)]
\(\Rightarrow \) a + c > 2b...(iii)
Similarly, for the last three terms
\(\frac { b+d }{ 2 } >c\)   [\(\because \sqrt { bd} =c\)]
\(\Rightarrow \) b + d > 2c....(iv)
On adding Eqs. (iii) and (iv), we get
(a + c) + (b + d) > 2b + 2c
\(\Rightarrow \) a + d > b + c

(i) Given a, b, c, d are in AP and we know that AM > GM, then for the first three terms.
b > \(\sqrt { ac } \)  [here, \(\frac { a+c }{ 2 } \) = b]
On squaring both sides, we get
b² > ac.....(i)
Similarly, for the last three terms
C > \(\sqrt { bd } \)   [here, \(\frac { b+d }{ 2 } \) = c]
\(\Rightarrow \)  c² > bd...(ii)
On multiplying Eqs. (i) and (ii), we get 
b² c² > (ac) (bd)
\(\Rightarrow \)  bc  > ad

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.

9T₉ = 13T₁₃  \(\Longrightarrow \) 9 ( a + 8d ) = 13 ( a + 12d ) \(\Longrightarrow \) a = -21d = 0