[tex]\huge \underline\frak \red{Answer}[/tex]
Consider a& d are the first & common difference. So, 4th term & 6th term is a+3d & a+5d respectively.Then,
a+3d= 24 & a+5d=96
Subtracting 4th term from 6th term,
or, 2d = 72
Therefore, d= 36
Putting the value of d= 36 in a+3d=24,then
a= 24–3d=24–3×36=24–108=(-84)
Now, we have,
a=(-84), d= 36, n=10
Sum of first 10 term,
S10=10/2[2×(-84)+(10–1)36] [S(n)=n/2{2a+(n-1)d}]
=5[-168+9×36]
=5[-168+324]
=5×256
=1280
