If the sum of the first n terms of two A.P.s are in the ratio (7n -5):

1.	If the sum of the first n terms of two A.P.s are in the ratio (7n -5):(5n + 17); show thathe 6th terms of the two progressions are equal.
Answer» Let us assume that the two APs are: a1 + (n-1)d1 And a2 + (n-1)d2 Now the sum of these respective APs are n/2(2a1 + (n-1)d1) n/2(2a2 + (n-1)d2) Taking the ratio we get (2a1 + (n-1)d1) : (2a2 + (n-1)d2) = (7n-5) : (5n+17) Comparing the coefficients of n on the RHS with those in both the terms on the LHS, we get d1= 7 and d2= 5 Substituting these values in the equation and comparing the constants we get, a1= 1 and a2= 11 Therefore the series are 1,8,15,22,29,36,43,50,57,64,71…. And 11,16,21,26,31,36,41,46,51,56,61,66,71,76… So the series of numbers common in both APs is 36,71,106,141,176,211,…. Or (35n+1) where n is a natural number. 35 because it's the LCM of 7 and 5, the common differences of the APs. So the number of equal terms are infinite.

If the sum of the first n terms of two A.P.s are in the ratio (7n -5):(5n + 17); show thathe 6th terms of the two progressions are equal.

Answer»

Let us assume that the two APs are:

a1 + (n-1)d1

And

a2 + (n-1)d2

Now the sum of these respective APs are

n/2(2*a1 + (n-1)d1)

n/2(2*a2 + (n-1)d2)

Taking the ratio we get

(2*a1 + (n-1)d1) : (2*a2 + (n-1)d2) = (7n-5) : (5n+17)

Comparing the coefficients of n on the RHS with those in both the terms on the LHS, we get

d1= 7 and d2= 5

Substituting these values in the equation and comparing the constants we get,

a1= 1 and a2= 11

Therefore the series are

1,8,15,22,29,36,43,50,57,64,71….

And

11,16,21,26,31,36,41,46,51,56,61,66,71,76…

So the series of numbers common in both APs is

36,71,106,141,176,211,….

Or (35n+1) where n is a natural number. 35 because it's the LCM of 7 and 5, the common differences of the APs.

So the number of equal terms are infinite.