Given {tex}a = 8, d =10 - 8 = 2{/tex}
{tex}\because{/tex} t_n = {tex}a + (n - 1 )d{/tex}
{tex}\therefore{/tex}t₆₀ = 8 + (60 - 1)2
= 8 + 59 {tex}\times{/tex} 2
= 8 + 118
= 126
t₅₁ = 8 + (51 -1) 2
= 8 + 50 {tex}\times{/tex} 2
= 8 + 100
= 108
Sum of last 10 terms = t₅₁ + t₅₂ + .... + t₆₀
Here,{tex} a = 108 {/tex} and {tex} d = 2{/tex}
Sn = {tex}\frac{n}{2}{/tex} {tex} [2a + (n-1)d]{/tex}
S₁₀ = {tex}\frac{10}{2}{/tex}[2{tex}\times{/tex}108 + (10 -1)2]
= 5[216 + 9(2)]
= 5(216 + 18)
= 5 {tex}\times{/tex} 234
{tex}=1170{/tex}
Hence, Sum of last 10 terms = 1170

Suppose man's 1 day's work be {tex}\frac 1x{/tex} and boy's 1 day's work be {tex}\frac 1y{/tex}
Suppose, {tex}\frac 1x{/tex}= u and {tex}\frac 1y{/tex}= v
By first condition,

{tex}\frac { 2 } { x } + \frac { 5 } { y } = \frac { 1 } { 4 } \Rightarrow 2 u + 5 v = \frac { 1 } { 4 }{/tex}...........(i)
By second condition,

{tex}\frac { 3 } { x } + \frac { 6 } { y } = \frac { 1 } { 3 } \Rightarrow 3 u + 6 v = \frac { 1 } { 3 }{/tex}.............(ii)
Multiplying (i) by 6 and (ii) by 5, 
{tex}\Rightarrow 12 u + 30 v = \frac { 6 } { 4 }{/tex}..........(iii)
{tex}15 u + 30 v = \frac { 5 } { 3 }{/tex}...........(iv)
Subtracting (iii) and (iv),
{tex}\Rightarrow 3u ={/tex} {tex} \frac { 5 } { 3 } - \frac { 6 } { 4 }{/tex}
{tex}\Rightarrow 3 u = \frac { 20 - 18 } { 12 }{/tex}
{tex}\Rightarrow 3 u = \frac { 2 } { 12 }{/tex}
{tex}\Rightarrow 3 u = \frac { 1 } { 6 }{/tex}
{tex}\Rightarrow u = \frac { 1 } { 18 }{/tex}
Putting u = {tex}\frac{1}{18}{/tex} in (i), 
{tex}\Rightarrow 2 \times \frac { 1 } { 18 } + 5 v = \frac { 1 } { 4 }{/tex}{tex}\Rightarrow \frac { 1 } { 9 } + 5 v = \frac { 1 } { 4 } \Rightarrow 5 v = \frac { 1 } { 4 } - \frac { 1 } { 9 }{/tex}
{tex}\Rightarrow 5 v = \frac { 5 } { 36 } \Rightarrow v = \frac { 1 } { 36 }{/tex}
Now, {tex}u ={/tex} {tex}\frac { 1 } { 18 } \Rightarrow x = \frac { 1 } { 4 } = 18{/tex}
and {tex}v ={/tex} {tex}\frac { 1 } { 36 } \Rightarrow y = \frac { 1 } { v } = 36{/tex}
{tex}\therefore{/tex}{tex} x = 18, y = 26{/tex}
The man will complete the given work in {tex}18\ days{/tex} and the boy will complete the given work in {tex}36 days{/tex} when they work alone.

A special economic zone is an area in which the business and trade laws are different from the rest of the country.

The given pair of equations are:
{tex}\frac{b}{a}x + \frac{a}{b}y = {a^2} + {b^2} {/tex}
So, {tex}\frac{b}{a}x + \frac{a}{b}y -[ {a^2} + {b^2} ] = 0{/tex} ...................(i)
And x + y = 2ab
x + y - 2ab = 0 ....................(ii)
Here,
{tex}{a_1} = \frac{b}{a},{b_1} = \frac{a}{b}{/tex}, c₁ = -(a² + b²)
a₂ = 1, b₂ = 1, c₂ = -(2ab)
By cross-multiplication method
{tex}\begin{array}{l}\;\frac x{{\displaystyle\frac ab}\times-(2ab)\;-1\lbrack-(a^2\;+\;b^2)\rbrack}=\;\frac y{-(a^2\;+\;b^2)\;-{\displaystyle\frac ba}\lbrack\;-(2ab)\rbrack}=\;\frac1{{\displaystyle\frac ba}-{\displaystyle\frac ab}}\\\frac x{\displaystyle\frac{-2a^2b}b\;\;+(a^2\;+\;b^2)}=\;\frac y{-(a^2\;+\;b^2)\;+{\displaystyle\frac{2ab^2}a}}=\;\frac1{\displaystyle\frac{b^2\;-\;a^2}{ab}\;}\\\end{array}{/tex}
{tex} \frac{x}{{{\frac ba - \frac ab} }} = \frac{{ - y}}{{ - {b^2} + {a^2}}} = \frac{1}{{\frac{{{b^2} - {a^2}}}{{ab}}}} {/tex}
{tex} \frac{x}{{{b^2} - {a^2}}} = \frac{{ - y}}{{ - {b^2} + {a^2}}} = \frac{1}{{\frac{{{b^2} - {a^2}}}{{ab}}}} {/tex}
{tex} \frac{x}{{{b^2} - {a^2}}} = \frac{1}{{\frac{{{b^2} - {a^2}}}{{ab}}}} {/tex}
{tex}⇒ x = ab{/tex}
And, {tex}\frac{{ - y}}{{ - {b^2} + {a^2}}} = \frac{1}{{\frac{{{b^2} - {a^2}}}{{ab}}}} {/tex}
{tex}⇒ y = ab{/tex}
The solutions of the given pair of equations is  x= ab and y = ab .

For (2p - 1)x + (p - 1 )y - (2p + 1) = 0

{tex}a _ { 1 } = 2 p - 1 , b _ { 1 } = p - 1 \text { and } c _ { 1 } = - ( 2 p + 1 ){/tex}

and for 3x + y - 1 = 0

{tex}a _ { 2 } = 3 , b _ { 2 } = 1 \text { and } c _ { 2 } = - 1{/tex}

The condition for no solution is

{tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }{/tex}

{tex}\frac { 2 p - 1 } { 3 } = \frac { p - 1 } { 1 } \neq \frac { 2 p + 1 } { 1 }{/tex}

By {tex}\frac { 2 p - 1 } { 3 } = \frac { p - 1 } { 1 }{/tex}

3/7-3 = 2 /7 -1

3/7 - 2/7 = 3 - 1

{tex}\therefore {/tex} p = 1

from {tex}\frac { p - 1 } { 1 } \neq 2 p + 1{/tex}

We have {tex}p - 1 \neq 2 p + 1 \text { or } 2 p - p = - 1 - 1{/tex}

{tex}p \neq - 2{/tex}

from {tex}\frac{{2p - 1}}{3} \ne \frac{{2p + 1}}{1}{/tex}

{tex}\Rightarrow \quad 2 p - 1 \neq 6 p + 3{/tex}

{tex} \Rightarrow \quad 4p \ne - 4{/tex}

{tex}p \neq - 1{/tex}

Here f(x)=x⁴ + 4x³ - 2x² - 20x - 15

{tex}\sqrt { 5 }{/tex}  and - {tex}\sqrt { 5 }{/tex} are zeros of f(x)
 {tex}\begin{array}{l}\text{so (x-}\sqrt5)(\mathrm x+\sqrt{5)}=\mathrm x^2-5\;\mathrm{is}\;\mathrm a\;\mathrm{factor}\;\mathrm{of}\;\mathrm f(\mathrm x)\\\end{array}{/tex}
Dividing x⁴ + 4x³ - 2x² - 20x - 15 by x²  - 5, we get

The quotient q(x)=x²+4x+3

=x²+x+3x+3

=x(x+1)+3(x+1)

=(x+3)(x+1)

q(x)=0 if x+3=0 or x+1 =0

Hence zeros of q(x) are -3 and -1
Thus, the zeros of the given polynomial f(x) are
{tex}\sqrt { 5 } , - \sqrt { 5 }{/tex} , -3 , -1