You can put this solution on YOUR website! 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
The problem does not define the meaning of S6, S4, so I will suppose that
S6 and S4 are the sums of the first 6 and the first 4 terms of the AP, respectively.
There are 13 - 6 = 7 gaps between the 13-th and 6-th terms of the AP on the number line,
so each gap is = 6 units.
Thus the common difference of the AP is d= 6.
Now, - = + = + = (35-6) + 35 = 70-6 = 64.
It is the answer to the problem's question: - = 64.