Question

please solve this problem​

Answer 1

Answer:

There are 6 terms in the progression.

Step-by-step explanation:

Let the first term be a, the common ratio be r and the number of terms be n.

Recall the k-th term is given by ar^(k-1).

We are given:

1. a + ar^(n-1) = 66
2. ar × ar^(n-2) = 128
3. a + ar + ... + ar^(n-1) = 126

Rearranging condition (2) gives:

    a × ar^(n-1) = 128

So the sum of a and ar^(n-1) is 66, and their product is 128.  This means these two numbers are the roots of the quadratic equation

  x² - 66x + 128 = 0

=> ( x - 2 ) ( x - 64 ) = 0

Since we're told the GP is increasing, it follows that

  a = 2

and ar^(n-1) = 64 => r^(n-1) = 64 / a = 64 / 2 = 32.

From condition (3),

a + ar + ... + ar^(n-1) = 126

=> a ( 1 + r + ... + r^(n-1) ) = 126

=> ( r^n - 1 ) / ( r - 1 ) = 126 / a = 126 / 2 = 63

=> r^n - 1 = 63 ( r - 1 )

Remember that r^(n-1) = 32, so continue...

=> 32r - 1 = 63 ( r - 1 )

=> 32r - 1 = 63r - 63

=> 31r = 62

=> r = 2

Finally, r^(n-1) = 32

=> 2^(n-1) = 32

=> n-1 = 5

=> n = 6

That is, there are 6 terms in the progression.