A man arranges to pay off …

Let first instalment(a) = Rs a
Let common difference = Rs d
Given,
Amount of 40 instalments = Rs 3600
{tex}\Rightarrow \frac { 40 } { 2 } [ 2 a + ( 40 - 1 ) d ] = 3600{/tex}
{tex}\Rightarrow{/tex}20[2a + 39d] = 3600
{tex}\Rightarrow{/tex}2a + 39d = 180..............(i)
and, Amount of 30 instalments = Rs 2400
{tex}\Rightarrow \frac { 30 } { 2 } [ 2 a + ( 30 - 1 ) d ] = 2400{/tex}
{tex}\Rightarrow{/tex}15[2a + 29d] = 2400
{tex}\Rightarrow{/tex}2a + 29d = 160..............(ii)
Subtracting (ii) from (i),
2a + 39d - 2a - 29d = 180 - 160
{tex}\Rightarrow{/tex} 10d = 20
{tex}\Rightarrow \quad d= \frac { 20 } { 10 } = 2{/tex}
Putting value of d in eq(i),
2a + 39{tex}\times{/tex} 2 = 180
{tex}\Rightarrow{/tex}2a + 78 = 180
{tex}\Rightarrow{/tex}2a = 180 - 78
{tex}\Rightarrow{/tex}2a = 102
{tex}\Rightarrow a = \frac { 102 } { 2 } = 51{/tex}
Therefore, Value of first instalment = Rs 51