Solution:

It is given that on the day first he made five frames then 2 frames more than the previous i.e. 7 and then 9 and so on

Therefore the making of frames each day forms a sequence of 5, 7, 9…

Sequence so formed is an AP with $a=2$ as the first term as and $d = 2$ as the common difference

The total no. of frames that is to be made $= 192$

Let’s assume that the required days is n hence ${{S}_{n}}~=\text{ }192$

Sum of the first n terms of the AP so formed is given by ${{S}_{n~}}=\text{ }n/2\text{ }\left( 2a\text{ }+\text{ }\left( n{-}1 \right)\text{ }d \right)$

Where the first term is ‘a’ and the common difference is ‘d’

${{S}_{n~}}=\text{ }\left( n/2 \right)\text{ }\left( 2\left( 5 \right)\text{ }+\text{ }\left( n{-}1 \right)\text{ }2 \right)$

$192\text{ }=\text{ }\left( n/2 \right)\text{ }\left( 10\text{ }+\text{ }2n{-}2 \right)$

$\Rightarrow ~384\text{ }=\text{ }10n\text{ }+\text{ }2{{n}^{2}}{-}2n$

When we compute and simplify,  we obtain

$\Rightarrow ~2{{n}^{2}}~+\text{ }8n{-}384\text{ }=\text{ }0$

$\Rightarrow ~{{n}^{2}}~+\text{ }4n{-}192\text{ }=\text{ }0$

$\Rightarrow ~{{n}^{2}}~+\text{ }16n{-}12n{-}192\text{ }=\text{ }0$

$\Rightarrow ~n\text{ }\left( n\text{ }+\text{ }16 \right){-}12\text{ }\left( n\text{ }+\text{ }16 \right)\text{}=\text{ }0$

$\Rightarrow ~(n{-}12)\text{ }\left( n\text{ }+\text{ }16 \right)\text{ }=\text{ }0$

$\Rightarrow ~n\text{ }=\text{ }12\text{ }and\text{ }n\text{ }={-}16$

But here we know that ‘n’ represents the no. of days and that cannot be negative therefore $n = 12$

As a result, 12 days is required to finish the job.