Question 933070
The sum of the first {{{n}}} terms is
{{{(2a+(n-1)*d)(n/2)}}} , so
the sum of the first {{{200}}} terms is
{{{(2a+(200-1)d)(200/2)=(2a+199d)(100)=200a+19900d}}} , and
the sum of the first {{{n}}} terms is
{{{(2a+(100-1)d)(100/2)=(2a+99d)(50)=100a+4950d}}} .
Since the sum of the first 200 terms is 4 times the sum of the first 100 terms,
{{{200a+19900d=4(100a+4950d)}}}
{{{200a+19900d=400a+19800d}}}
{{{100d=200a}}}
{{{d=200a/100}}}
{{{highlight(d=2a)}}}
The {{{n^th}}} therm is {{{a+(n-1)*d}}} , so the {{{100^th}}} term is
{{{a+(n-1)*(2a)=a(1+(n-1)*2)=a(1+2n-2)=highlight(a(2n-1))}}} .