The formula for the nth term of an AP is: xn=a+(n−1)d
x5=a+(5−1)d=a+4d
Since x5=−4:
a+4d=−4 ... (equation 1)
Given: 2x6+2x9=x11+x13
2(x6+x9)=x11+x13
x6=a+5d
x9=a+8d
x11=a+10d
x13=a+12d
Substituting:
x6+x9=(a+5d)+(a+8d)=2a+13d
x11+x13=(a+10d)+(a+12d)=2a+22d
Our equation becomes:
2(2a+13d)=2a+22d
4a+26d=2a+22d
From our simplified equation:
4a+26d=2a+22d
4a−2a=22d−26d
2a=−4d
a+2d=0 ... (equation 2)
Now we have two equations:
Equation 1: a+4d=−4
Equation 2: a+2d=0
Subtracting equation 2 from equation 1:
(a+4d)−(a+2d)=−4−0
a+4d−a−2d=−4
2d=−4
d=−2
Finding a: Substitute d=−2 into equation 2:
a+2(−2)=0
a−4=0
a=4
Using our AP formula with a=4 and d=−2:
x100=a+(100−1)d
=4+99(−2)
=4−198
=−194