Here `a=100`. Let the common diference be `d`.
The sum of the first six terms `t_(1)+t_(2)+…………+t_(6)`
`=a+(a+d)+...+(a+5d)`.
The sum of the six terms `=t_(7)+t_(8)+….+t_(12)`
`=(a+6d)+(a+8d)+….+(a+11d)`
From the given condition,
`(t_(1)+t_(2)+t_(3)+.........+t_(6))=5(t_(7)+t_(8)+.....+t_(12))`
`[(a+(a+d)+.............+a+5d)]=5[(a+6d)+(a+7d)+...........+(a+11d)]`
`:.[6a+(1+2+....+5)d]=[(6a+6+7+.............+11)d]`.............1
Now `1+2+.......+5=n/2(t_(1)+t_(n))=5/2(1+5)=5/2(1+5)=5/2xx6=15`.........2
and `6+7+............+11=n/2(t_(1)+t_(n))=6/2(6+11)=3xx17=51`.............3
From 1, 2 and 3
`6a+15d=(6a+51d)`
`:.6a+15d=30a+255d`
`:.30a-6a=-255d+15d`
`:.24a=-240d`
`:.a=-10d`........(Dividing both the sides by 24)
`:.100=-10d` ........(Substituting `a=100`) .........(Given)
`:.d=-10`
Now `a=-t_(1)=100,d=-10`
`:.t_(2)+t_(1)+d=100-10=90,t_(3)=t_(2)+d=90-10=80`
Ans. The required A.P. is 100, 90, 70............