- the first AP 1,3,5,...,1991 have a= 1 d=2 tn = 1991
- the second ap 1,6,11,...,1991 have a= 1 d= 5 tn= 1991

the first common term is 11 in ap no. 1 it is 6 th term and in ap no. 2 it is 3 rd term

the common term repeat after 2 terms in ap no. 2

so, here a= 11 d=10

last term of both term are same

so, tn=1991

tn = a+(n-1)d

1991=11+(n-1)10

1980=(n-1)10

198=n-1

so,

n=199

sn=n/2[a+tn]

=199/2*[11+1991]

=199/2*[2002]

=199/2*2002

=199*1001

so,

sn=199199

hope this will helpful

sorry i am bad in explanation part.

i just find the first common term and of both ap and divide them and found the common difference that is 10.

i was unable to find proper reason for steps but you could understand how to do such sums.

sorry for incontinence and thank you for such a nice sum :)