Solve the following
Question:

Solve:

$\frac{2-9 z}{17-4 z}=\frac{4}{5}$

Solution:

$\frac{2-9 z}{17-4 z}=\frac{4}{5}$

$\Rightarrow 5(2-9 z)=4(17-4 z) \quad$ (by cross multiplication)

$\Rightarrow 10-45 z=68-16 z$

$\Rightarrow 10-68=45 z-16 z$

$\Rightarrow-58=29 z$

$\Rightarrow 29 z=-58 \quad$ (by transposition)

$\Rightarrow z=\frac{-58}{29}=-2$

$\therefore z=-2$

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Solve the following
Question:

72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.

Solution:

Let P(n) be the given statement.

Now,

$P(n): 7^{2 n}+2^{3 n-3} \cdot 3^{n-1}$ is divisible by 25

Step 1:

$P(1): 7^{2}+2^{3-3} \cdot 3^{1-1}=49+1=50$

It is divisible by 25 .

Thus, $P(1)$ is true

Step 2: Let $P(m)$ be true.

Now,

$7^{2 m}+2^{3 m-3} \cdot 3^{m-1}$ is divisible by 25 .

Suppose :

$7^{2 m}+2^{3 m-3} \cdot 3^{m-1}=25 \lambda \quad \ldots(1)$

We have to show that $P(m+1)$ is true whenever $P(m)$ is true.

Now,

$P(m+1)=7^{2 m+2}+2^{3 m} \cdot 3^{m}$

$=7^{2 m+2}+7^{2} \cdot 2^{3 m-3} \cdot 3^{m-1}-7^{2} \cdot 2^{3 m-3} \cdot 3^{m-1}+2^{3 m} \cdot 3^{m}$

$=7^{2}\left(7^{2 m}+2^{3 m-3} \cdot 3^{m-1}\right)+2^{3 m} \cdot 3^{m}\left(1-\frac{49}{24}\right)$

$=7^{2} \times 25 \lambda-2^{3 m} \cdot 3^{m} \times \frac{25}{2^{3} \cdot 3^{1}} \quad[$ Using $(1)]$

$=25\left(49 \lambda-2^{3 m-3} \cdot 3^{m-1}\right)$

It is divisible by 25 .

Thus, $P(m+1)$ is true.

By the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.

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