Question

Find absolute value of the difference of the arithmetic progression an if a1+a2=5 and a1^+a2^2=13

Answer 1

Arithmetic progression

Given, $a_{1}^{2}+a_{2}^{2}=13$

$\Rightarrow (a_{1}+a_{2})^{2}-2a_{1}a_{2}=13$

$\Rightarrow 5^{2}-2a_{1}a_{2}=13\quad[\because a_{1}+a_{2}=5]$

$\Rightarrow 25-2a_{1}a_{2}=13$

$\Rightarrow 2a_{1}a_{2}=12$

$\Rightarrow 4a_{1}a_{2}=24$

Now, $a_{1}+a_{2}=5$

$\Rightarrow (a_{1}+a_{2})^{2}=5^{2}$

$\Rightarrow (a_{1}-a_{2})^{2}+4a_{1}a_{2}=25$

$\Rightarrow (a_{1}-a_{2})^{2}+24=25\quad[\because 4a_{1}a_{2}=24]$

$\Rightarrow (a_{1}-a_{2})^{2}=1$

$\Rightarrow a_{1}-a_{2}=\pm 1$

$\Rightarrow \color{blue}{|a_{1}-a_{2}|=1}$

Answer: Hence, absolute value of the common difference is $\bold{1}$.

Answer 2

Given:

$a+b=5\\a^2+b^2=13$

To Find:

|a-b|=?

Solution:

Formula:

$(a+b)^2-(a^2+b^2)=2ab\\\\(a-b)^2=(a^2+b^2)-2ab\\\\a-b= \sqrt{(a^2+b^2)-2ab}$

Find the value of ab:

$\to 5^2-13=2ab\\\\\to 25-13=2ab\\\\\to 12 =2ab\\\\\to 2ab= 12\\\\\to ab=6$

find a-b:

$\to a-b=\sqrt{(a^2+b^2) -2ab}\\\\\to a-b=\sqrt{(13) -2\times 6}\\\\\to a-b=\sqrt{13 -12}\\\\\to a-b=\sqrt{1}\\\\\boxed {\to a-b= 1}$

The final value is 1.