Question 858287
two arithmetic series are such that their common difference are 9 and 3 respectively. If their first terms are 2 and 5 respectively, find the number of terms of each series that would give a common sum. 


Let Sn be the sum of first series d=9, a=2

S'n be the sum of second series. d=3, a=5

Sn = n/2(2a1+(n-1)d)
Sn = n/2 (2*2+(n-1) 9)

Sn = n/2( 4+9n-9)
Sn = n/2(4n-5)

Find S'n

S'n = n/2(2*5+(n-1)*3)

S'n = n/2(10+3n-3)

S'n = n/2(3n+7)

Sn=S'n

n/2(4n-5)=n/2(3n+7)

4n-5 = 3n+7

4n-3n=12
n=12

For the 12 th. term the sums are equal