# What is the sum of all the terms common to the artithmetic progressions 1,3,5....., 1991 and 1,6,11, ....., 1991?

Sn=[ n/2 {2a+ (n-1)d}]

or

Sn=[a+l]d
• -2
1. the first AP        1,3,5,...,1991 have a= 1 d=2 tn = 1991
2. 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*
=199/2*2002
=199*1001
so,
sn=199199

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 :)

• -1
Infinity
• -1
Sn= n/2 ( 2 a + (n-1)?d • -2
You are good in explanation
• 0
What are you looking for?