Question

Solve the question attached

Class 11th
Maths
Harmonic Progression

Right answer will get brainiest
Wrong will straight away reported ​

Answer 1

Question :

If $\bf\:x_{1},x_{2},x_{3},x_{4},x_{5}$ are in HP

prove that $\bf\:x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{4}+x_{4}x_{5}=4x_{1}x_{5}$

Theory :

If a,b,c are in HP then

$\sf\:\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in AP

Solution :

$\sf\:x_{1},x_{2},x_{3},x_{4},x_{5}$ are in HP

Therefore

$\sf\:\dfrac{1}{x_{1}},\dfrac{1}{x_{2}},\dfrac{1}{x_{3}},\dfrac{1}{x_{4}},\dfrac{1}{x_{5}}$ are in AP

Let the common difference of AP be d

$\sf\:d=\dfrac{1}{x _{2}}-\dfrac{1}{x _{1}}=\dfrac{x_{1}-x_{2}}{x _{2}x_{1}}$

$\sf\:\implies\:x_{1}x_{2}=\dfrac{x_{1}-x_{2}}{d}$

Similarly,

$\sf\:d=\dfrac{1}{x _{3}}-\dfrac{1}{x _{2}}=\dfrac{x_{2}-x_{3}}{x _{3}x_{2}}$

$\sf\:\implies\:x_{3}x_{2}=\dfrac{x_{2}-x_{3}}{d}$

and

$\sf\:d=\dfrac{1}{x _{5}}-\dfrac{1}{x _{4}}=\dfrac{x_{4}-x_{5}}{x _{4}x_{5}}$

$\sf\:\implies\:x_{4}x_{5}=\dfrac{x_{4}-x_{5}}{d}$

we know that ,

Genral term of an ap=$\sf\:T_{n}=a+(n-1)d$

Therefore,nth term of the AP series .

$\sf \frac{1}{x_{5}} = \frac{1}{x_{1}} + (n - 1)d$

$\implies \sf \dfrac{x_{1} - x_{5} }{ x_{4} x_{5} } = 4d$

$\sf\:\implies\:\dfrac{x_{1} - x_{5}}{4x_{4} x_{5}}=d...(1)$

___________________________

We have to prove that $\sf\:x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{4}+x_{4}x_{5}=4x_{1}x_{5}$

LHS= $\sf\:x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{4}+x_{4}x_{5}$

$\sf\:=\dfrac{1}{d}(x_{1}- x_{2}+x_{2}- x_{3}+x_{3} -x_{4}+x_{4} +x_{5})$

$\sf\:=\dfrac{1}{d}(x_{1}- x_{5})$

Now put the value of d from (1),then we get

$\sf\:=4x_{1}x_{5}$

=RHS

⇒LHS =RHS

Hence proved

Answer 2

Solve: (2x + 5)/(x + 4) = 1

Solution:

(2x + 5)/(x + 4) = 1

⇒ 2x + 5 = 1(x + 4)

⇒ 2x + 5 = x + 4

⇒ 2x - x = 4 - 5 (Transferring positive x to the left hand side changes to negative x and again, positive 5 changes to negative 5)

⇒ x = -1

Therefore, x = - 1 is the required solution of the equation (2x + 5)/(x + 4) = 1

Solution:

6x - 19 = 3x - 10

⇒ 6x - 3x = - 10 + 19 (Transferring 3x to L.H.S changes to negative 3x and -19 to R.H.S. changes to positive 19)

⇒ 3x = 9

⇒ 3x/3 = 9/3 (Dividing both sides by 3)

⇒ x = 3

3. Solve: 5 - 2(x - 1) = 4(3 - x) - 2x.

Solution:

5 - 2(x - 1) = 4(3 - x) - 2x

⇒ 5 - 2x + 2 = 12 - 4x - 2x (Removing the brackets and then simplify)

⇒ 7 - 2x = 12 - 6x (Transferring -6x to L.H.S. changes to positive 6x and 7 to R.H.S. changes to negative 7)

⇒ -2x + 6x = 12 - 7

⇒ 4x = 5

⇒ 4x/4 = 5/4

⇒ x = 5/4

4. x/2 + x/3 = x - 7

Solution:

x/2 + x/3 = x - 7

Least common multiple of2 and 3 is 6

⇒ (x/2 × 3/3) + (x/3 × 2/2) = x - 7

⇒ 3x/6 + 2x/6 = x - 7

⇒ (3x + 2x)/6 = x - 7

⇒ 5x/6 = x - 7

⇒ 5x = 6(x - 7)

⇒ 5x = 6x - 42 (Transferring 6x to L.H.S. changes to negative 6x)

⇒ 5x - 6x = -42

⇒ -x = -42

⇒ x = 42

5. 2/(x + 3) = 3/(5 - x)

Solution:

2/(x + 3) = 3/(5 - x)

⇒ 3(x + 3) = 2(5 - x) (cross multiply and then remove the brackets)

⇒ 3x + 9 = 10 - 2x (Transferring -2x to L.H.S. changes to positive 2x and 9 to R.H.S. changes to -9)

⇒ 3x + 2x = 10 - 9

⇒ 5x = 1

⇒ 5x/5 = 1/5 (Dividing both sides by 5)

⇒ x = 1/5