# If A and G be A.M and G.M, respectively between two positive numbers, prove that the numbers are A ± √( A + G)( A - G

**Solution:**

It is given that A and G are A.M and G.M, respectively between two positive numbers.

Let these two positive numbers be a and b.

Therefore,

A.M = A = (a + b) / 2

⇒ a + b = 2 A ....(1)

And,

G.M = G = √ab

⇒ ab = G^{2} ....(2)

Since,

(a - b)^{2} = (a + b)^{2} - 4ab

= 4 A^{2} - 4G^{2} [Using (1) and (2)]

= 4( A^{2} - G^{2})

= 4( A + G)( A - G)

(a - b) = 2√( A + G)( A - G) ....(3)

By adding (1) and (3), we obtain

⇒ 2a = 2A + 2√( A + G)( A - G)

⇒ a = A + √( A + G)( A - G)

Substituting the value of a in (1) , we obtain

b = 2 A - (A + √( A + G)( A - G))

= A - √( A + G)( A - G))

Thus, the two numbers are A ± √( A + G)( A - G))

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 29

## If A and G be A.M and G.M, respectively between two positive numbers, prove that the numbers are A ± √( A + G)( A - G)

**Summary:**

A and G are the A.M and G.M, respectively between two positive numbers we have proved that the numbers are A ± √( A + G)( A - G))