Question 1178050
Given t6=35
t13=77

therefore, t6= a+(n-1)d=a+(6-1)d
35=a+5d .....1

t13= a+(13-1)d= a+12d
77=a+12d ...2

by solving 1 and 2 we have
a=5, d=6

now, sn=n/2(2a+(n-1)d) using this formula
s6=6/2(2*5+(6-1)*6)
s6=3(10+5*6)
s6=120

similary,s4=4/2(2*5+(4-1)*6)
s4=2(10+3*6)
s4=56

therefore, s6-s4= 120-56
s6-s4=64