Question

plzzz ans....Thnk u....Plzzz ans the first part

Answer 1

since , a : b :: b : c

=> a/b = b/c => b² = ac

=> a , b , c are in geometric progression

In a geometric progression , the number is increased by a nth power of common ratio

=> a , ar , ar² , ar³ , ar⁴ .......... ar^nth

where r is any arbitrary number other than 0 and 1

if we take => a = a

=> b = ar

=> c = ar²
and divide it by 1/r

We get ,

a = a / r

b = a

c = ar

so , now in LHS

=> ( a+b ) / ( b+c )

=> ( a/r + a ) / ( a + ar )

=> [ (a + ar) / r ] / ( a+ar )

=> (a+ar) gets cancelled out

so the result is 1 / r


Now , RHS => a² ( b-c ) / b² ( a-b )

=> ( a/r )² ( a-ar ) / a² ( a/r - a )

=> a² / r² × a ( 1 - r ) / a² × a ( 1/r - 1 )

=> [ a³ ( 1-r ) / r² ] / a³( 1-r ) / r

=> a³( 1 - r ) gets cancelled out

=> ( 1 / r² ) / ( 1 / r )

=> r / r² => 1 / r


since , LHS = RHS

hence , Proved


hope this helps

Answer 2

a+b/b+c = a²(b-c) /b²(a-b)
=> (a+b)(a-b) b² = a²(b-c) (b+c)
=> ( a²-b²) b² = a² ( b²-c²)
=> a²b²-b⁴= a²b²-a²c²
=> a²c² =b⁴
=> ac = b².

We got ac = b² .

If a, b, c are in continued proportion,
a : b :: b:c then b²=ac.

Hence, proved that if a, b, c are in proportion .

Also, we can prove it in other way.
We know that for a, b, c in continued proportion ac=b²

Squaring on both sides .
a²c²=b⁴
Add -a²b² on both sides.
a²c²-a²b² = b⁴-a²b²
a²(c²-b²) = b²(b²-a²)
a²(c+b)(c-b) = b²(b+a)(b-a)

Multiplying by -1 on both sides.

-a²(c+b)(c-b) = -b²(b+a)(b-a)
a²(c+b)(b-c) = b²(b+a)(a-b)

Transpose terms now,
a+b/b+c = a²(b-c)/b²(a-b)

Hope helped!