A  body  executes S.H.M. under the action of a force F₁ with a time period 7/6  seconds.  If the force is changed to F₂ it executes S.H.M. with time period 7/8  seconds. If both the forces F₁ and F₂ act simultaneously in the same direction on  the body, then its time period in seconds is 

 For a body executing simple harmonic motion, the force can be written as $F=m{\omega }^{2}y$  
For the first force $F_1=m{\left(\frac{2\pi }{T_1}\right)}^{2}y$  and for the second force  $F_2=m{\left(\frac{2\pi }{T_2}\right)}^{2}y$. 
On combining the two forces  $F_1+F_2=F_{net}$ , where $F_{net}=m{\left(\frac{2\pi }{T_{net}}\right)}^{2}y$ 
Substituting this in the equation $F_1+F_2=F_{net}$, we get
$m{\left(\frac{2\pi }{T_{net}}\right)}^{2}y=m{\left(\frac{2\pi }{T_1}\right)}^{2}y+m{\left(\frac{2\pi }{T_2}\right)}^{2}y$ 
From the question,  $T_1=\frac{7}{6}$ s and  $T_2=\frac{7}{8}$ s
$\begin{array}{l}{\left(\frac{1}{T_{net}}\right)}^2={\left(\frac{6}{7}\right)}^2+{\left(\frac{8}{7}\right)}^2\\ {\left(\frac{1}{T_{net}}\right)}^2=\left(\frac{100}{49}\right)\\ T_{net}^2=\left(\frac{49}{100}\right)\end{array}$ 
Therefore,  $T_{net}=\left(\frac{7}{10}\right)s$