A  body  executes S.H.M. under the action of a force $F_1$ with a time period 7/6  seconds.  If the force is changed to $F_2$ it executes S.H.M. with time period 7/8  seconds. If both the forces $F_1$ and $F_2$ 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 acting on the body= $F=m{\omega }^{2}y$  
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 these in the equation $F_1+F_2=F_{net}$,

 we get
 $m{\left(\frac{2\pi }{T_{net}}\right)}^{2}y=m{\left(\frac{2\pi }{T_1}\right)}^{2}y+m{\left(\frac{2\pi }{T_2}\right)}^{2}y$
by the question,  $T_1=\frac{7}{6}$ s and $T_2=\frac{7}{8}$  s
 ${\left(\frac{1}{T_{net}}\right)}^2={\left(\frac{6}{7}\right)}^2+{\left(\frac{8}{7}\right)}^2$
${\left(\frac{1}{T_{net}}\right)}^2=\left(\frac{100}{49}\right)$ 
 ${T_{net}}^2=\left(\frac{49}{100}\right)$
Therefore,  $T_{net}=\left(\frac{7}{10}\right)s$