Question

Two AP contains the same no. of terms. The ratio of the last term of the first AP to the first term of second AP is equal to the ratio of the last term of second AP to the first term of the first AP and is equal to 4.
The ratio of the sum of the first AP to that of the second AP is 2. Find the ratio of the differences of the progressions.

Answer 1

Let the terms be n:

So they contain the same terms: n

Last term of first AP : \(a_n = a_x + (n-1)d_x\)

So:

\[\frac{a_x+ (n-1)d_x}{a_y} = \frac{a_y + (n-1)d_y}{a_x} = 4\]

\[\large \frac{\frac{n}{2}(2 a_x + (n-1)d_x)}{\frac{n}{2}(2 a_y + (n-1)d_y)} = 2\]

Answer 2

\(a_x\) : first term of first AP

\(a_y\) : first term of second AP.

Answer 3

ok

Answer 4

now

Answer 5

?

Answer 6

Wait let me find easier method to solve this..

Answer 7

This has lengthy calculations..

But still I am close to it..

Answer 8

ok

Answer 9

I think today is not my day..

Answer 10

@mukushla can help.

Answer 11

\[a_x + (n-1)d_x = 4a_y\]
\[a_y + (n-1)d_y = 4a_x\]

Answer 12

\[\frac{a_x + 4a_y}{a_y + 4a_x} = 2 \implies \frac{a_x}{a_y} = \frac{2}{7}\]

Answer 13

now?

Answer 14

Got it I think..

Answer 15

Find \((n-1)\) from the two equations...

Answer 16

\[n-1 = \frac{4a_y - a_x}{d_x}\]

\[n - 1 = \frac{4a_x - a_y}{d_y}\]

Answer 17

Equate them..

Answer 18

\[\frac{d_x}{d_y} = \frac{4a_y - a_x}{4a_x - a_y}\]

And you know : \(a_x = \frac{2}{7}a_y\)

Try to go further..