If $a_1,a_2,a_3,a_4$  are the coefficients of 2^nd ,3^rd, 4^th and 5^th terms of ${\left(1+x\right)}^n$  respectively then $\frac{a_1}{a_1+a_2},\frac{a_2}{a_2+a_3},\frac{a_3}{a_3+a_4}$  are in

$t_1=\frac{a_1}{a_1+a_2}=\frac{{}^n{C}_1}{{}^n{C}_1+{}^n{C}_2}=\frac{n}{n+\frac{n\left(n-1\right)}{2}}=\frac{2n}{2n+n^2-n}$

$=\frac{2n}{n^2+n}=\frac{2}{n+1}$

$\frac{a_2}{a_2+a_3}=\frac{\frac{n\left(n-1\right)}{2}}{\frac{n\left(n-1\right)}{2}+\frac{n\left(n-1\right)\left(n-2\right)}{6}}$

$=\frac{\frac{n\left(n-1\right)}{2}}{\frac{3n\left(n-1\right)+n\left(n-1\right)\left(n-2\right)}{6}}$

$=\frac{n\left(n-1\right)}{2}\times \frac{6}{n\left(n-1\right)\left(3+n-2\right)}$

$=\left(\frac{3}{n+1}\right)$

$t_3=\frac{a_3}{a_3+a_4}=\frac{\frac{n\left(n-1\right)\left(n-2\right)}{6}}{\frac{n\left(n-1\right)\left(n-2\right)}{6}+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{24}}$

$=\frac{\left(\frac{n\left(n-1\right)\left(n-2\right)}{6}\right)}{\frac{n\left(n-1\right)\left(n-2\right)\left[4+\left(n-3\right)\right]}{24}}$

$=\frac{1}{6}\times \frac{24}{n+1}$

$=\frac{4}{n+1}$

Clearly $t_2-t_1=t_3-t_2$

$t_1,t_2,t_3$  are in A.P.