Find the common difference of an AP, whose first term is 100 and the sum of whose first six terms is five times the sum of the next six terms.

Explanation

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............