Answer :

It is given that,

$f\left(x+y\right)=f\left(x\right) \times f\left(y\right)$ for all $x, y \in N\qquad \ldots\left(1\right)$

$f\left(1\right)=3$

Taking $x=y=1$ in $\left(1\right)$, we obtain

$f\left(1+1\right)=f\left(2\right)=f\left(1\right)\times f\left(1\right)=3 \times 3=9$

Similarly,

$f\left(1+1+1\right)=f\left(3\right)=f\left(1+2\right)=f\left(1\right) f\left(2\right)=3 \times 9=27$

$f\left(4\right)=f\left(1+3\right)=f\left(1\right) f\left(3\right)=3 \times 27=81$

$\therefore \ \ f\left(1\right), f\left(2\right), f\left(3\right), \ldots$, that is $3,9,27, \ldots$, forms a $G.P.$ with both the first term and common ratio equal to $3 .$

It is known that,

$ S_n=\dfrac{a\left(r^{n}-1\right)}{r-1} $

It is given that, $\sum _ {x=1}^{n} f\left(x\right)=120$

$\therefore \ \ 120=\dfrac{3\left(3^{n}-1\right)}{3-1}$

$\Rightarrow 120=\dfrac{3}{2}\left(3^{n}-1\right)$

$\Rightarrow 3^{n}-1=80$

$\Rightarrow 3^{n}=81=3^{4}$

$\therefore \ \ n=4$

Thus, the value of $n$ is $4 .$

Answer :

Let the sum of $n$ terms of the $G.P.$ be $315 .$

It is known that, $S_n=\dfrac{a\left(r^{n}-1\right)}{r-1}$

It is given that the first term $a$ is $5$ and common ratio $r$ is $2 .$

$\therefore \ \ 315=\dfrac{5\left(2^{n}-1\right)}{2-1}$

$\Rightarrow 2^{n}-1=63$

$\Rightarrow 2^{n}=64=\left(2\right)^{6}$

$\Rightarrow n=6$

$\therefore \ \ $ Last term of the G.P $=6^{\text{th }}$ term $=a r^{6 - 1}=\left(5\right)\left(2\right)^{5}=\left(5\right)\left(32\right)=160$

Thus, the last term of the $G.P.$ is $160.$

Answer :

Let $a$ and $r$ be the first term and the common ratio of the $G.P.$ respectively.

$\therefore \quad a_{1}=1$

$a_3=a r^{2}=r^{2}$

$a_5=a r^{4}=r^{4}$

$\therefore \ \ r^{2}+r^{4}=90$

$\Rightarrow r^{4}+r^{2}- 90=0$

On solving the equation

$\Rightarrow r^{2}=\dfrac{-1+\sqrt{1+360}}{2}=\dfrac{-1 \pm \sqrt{361}}{2}$

$\Rightarrow \dfrac{-1 \pm 19}{2}=-10$ or $9$

$\therefore \ \ r= \pm 3 \ \ $ (Taking real roots)

Thus, the common ratio of the $G.P.$ is $\pm 3$.

Answer :

Let the three numbers in $G.P.$ be $a$, $a r$, and $a r^{2}$.

From the given condition, $a+a r+a r^{2}=56$

$\Rightarrow a\left(1+r+r^{2}\right)=56$

$\Rightarrow a=\dfrac{56}{1+r+r^{2}}\qquad \ldots\left(1\right)$

$a - 1 , ar -$ $7, a r^{2} - 21$ forms an $A.P.$

$ \therefore \ \ \left(ar - 7\right)-\left(a-1\right) = \left(ar^2 - 21\right) - \left(ar - 7\right) $

$\Rightarrow ar - a - 6=a r^{2} - ar - 14$

$\Rightarrow a r^{2} - 2 a r+a=8$

$\Rightarrow a r^{2}- a r- a r+a=8$

$\Rightarrow a\left(r^{2}+1 - 2 r\right)=8$

$\Rightarrow a\left(r- 1\right)^{2}=8$

$\Rightarrow \dfrac{56}{1+r+r^{2}}\left(r-1\right)^{2}=8$

[Using $\left(1\right)$]

$\Rightarrow 7\left(r^{2} - 2 r+1\right)=1+r+r^{2}$

$\Rightarrow 7 r^{2} - 14 r + 7 - 1 - r - r^{2}=0$

$\Rightarrow 6 r^{2} - 15 r+6=0$

$\Rightarrow 6 r^{2} - 12 r - 3 r+6=0$

$ \therefore \ \ {6r\left(r-2\right)-3\left(r-2\right)} $

$\Rightarrow\left(6 r- 3\right)\left(r- 2\right)=0$

$\therefore \ \ r=2, \dfrac{1}{2}$

From $\left(1\right)$

When $r=2, a=8$

When $r=\dfrac{1}{2}, a=32$

Therefore, when $r=2$, the three numbers in $G.P.$ are $8, 16,$ and $32.$

When $r=\dfrac{1}{2}$, the three numbers in $G.P.$ are $32, 16,$ and $8.$

Thus, in either case, the three required numbers are $8,16 ,$ and $32 .$

Answer :

Let the $G.P.$ be $T_1, T_2, T_3, T_4, \ldots T _ {2 n}$.

Number of terms $=2 n$

According to the given condition,

$T_1+T_2+T_3+\ldots+T _ {2 n}=5[T_1+T_3+\ldots+T _ {2_n{-1}}]$

$\Rightarrow T_1+T_2+T_3+ \ldots +T_{2 n}- 5[T_1+T_3+\ldots + T_{2n-1}] =0$

$\Rightarrow T_2+T_4+\ldots+T _ {2 n}=4[T_1+T_3+\ldots+T _ {2 n {-1}}]$

Let the $G.P.$ be $a, a r, a r^{2}, a r^{3}, \ldots$

$\therefore \ \ \dfrac{ar\left(r^{n}-1\right)}{r-1}=\dfrac{4 \times a\left(r^{n}-1\right)}{r-1}$

$\Rightarrow a r=4 a$

$\Rightarrow r=4$

Thus, the common ratio of the $G.P.$ is $4.$

Answer:

It is given that,

$ \dfrac{a+b x}{a-b x}=\dfrac{b+c x}{b-c x} $

$ \Rightarrow\left(a+b x\right)\left(b-c x\right)=\left(b+c x\right)\left(a-b x\right) $

$ \Rightarrow a b-a c x+b^{2} x-b c x^{2}=a b-b^{2} x+a c x-b c x^{2} $

$ \Rightarrow 2 b^{2} x=2 a c x $

$ \Rightarrow b^{2}=a c $

$ \Rightarrow \dfrac{b}{a}=\dfrac{c}{b} \qquad \ldots{\left(1\right)}$

Also, $\dfrac{b+c x}{b-c x}=\dfrac{c+d x}{c-d x}$

$\Rightarrow\left(b+c x\right)\left(c-d x\right)=\left(b-c x\right)\left(c+d x\right)$

$\Rightarrow b c-b d x+c^{2} x-c d x^{2}=b c+b d x-c^{2} x-c d x^{2}$

$\Rightarrow 2 c^{2} x=2 b d x$

$\Rightarrow c^{2}=b d$

$\Rightarrow \dfrac{c}{b}=\dfrac{d}{c}\qquad \ldots\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we obtain

$ \dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{c} $

Thus, $a, b, c$, and $d$ are in $G.P.$

Answer :

Let the $G.P.$ be $a, a r, a r^{2}, a r^{3}, \ldots a r^{n-1} \ldots$

According to the given information,

$ \begin{aligned} & S =\dfrac{a\left(r^{n}-1\right)}{r-1} \\ \\ & P=a^{n} \times r^{1+2+\ldots+n-1} \\ \\ & \quad =a^{n} r^{\dfrac{n\left(n-1\right)}{2}} \qquad {[\because \text{ Sum of first } n \text{ natural numbers is } n \dfrac{\left(n+1\right)}{2}]} \\ \\ &\therefore \ \ P^2=a^{2n}r^{n\left(n-1\right)} \\ \\ & R=\dfrac{1}{a}+\dfrac{1}{a r}+\ldots+\dfrac{1}{a r^{n-1}} \\ \\ & \quad =\dfrac{r^{n-1}+r^{n-2}+\ldots r+1}{a r^{n-1}} \\ \\ & \quad =\dfrac{1\left(r^{n}-1\right)}{\left(r-1\right)} \times \dfrac{1}{a r^{n-1}} \quad[\because 1, r, \ldots r^{n-1} \text{ forms a G.P }] \\ \\ & \quad =\dfrac{r^{n}-1}{a r^{n-1}\left(r-1\right)} \\ \\ & \therefore \ \ P^{2} R^{n}=a^{2 n} r^{n\left(n-1\right)} \dfrac{\left(r^{n}-1\right)^{n}}{a^{n} r^{n\left(n-1\right)}\left(r-1\right)^{n}} \\ \\ & \qquad\qquad=\dfrac{a^{n}\left(r^{n}-1\right)^{n}}{\left(r-1\right)^{n}} \\ \\ &\qquad\qquad =[\dfrac{a\left(r^{n}-1\right)}{\left(r-1\right)}]^{n} =S^{n} \end{aligned} $

Hence, $ \ \ P^{2} R^{n}=S^{n}$

Answer :

It is given that $a, b, c$, and $d$ are in $G.P.$

$\therefore \ \ b^{2}=a c\qquad \ldots\left(1\right)$

$c^{2}=b d\qquad \ldots\left(2\right)$

$a d=b c\qquad \ldots\left(3\right)$

It has to be proved that $\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right),\left(c^{n}+d^{n}\right)$ are in $G.P.$ i.e.,

$\left(b^{n}+c^{n}\right)^{2}=\left(a^{n}+b^{n}\right)\left(c^{n}+d^{n}\right)$

Consider $\text{L.H.S.}$

$\left(b^{n}+c^{n}\right)^{2}=b^{2 n}+2 b^{n} c^{n}+c^{2 n}$

$\qquad\qquad \ =\left(b^{2}\right)^{n}+2 b^{n} c^{n}+\left(c^{2}\right)^{n}$

$\qquad\qquad \ =\left(a c\right)^{n}+2 b^{n} c^{n}+\left(b d\right)^{n}[ $ Using $\left(1\right)$ and $\left(2\right)]$

$\qquad\qquad \ =a^{n} c^{n}+b^{n} c^{n}+b^{n} c^{n}+b^{n} d^{n}$

$\qquad\qquad \ =a^{n} c^{n}+b^{n} c^{n}+a^{n} d^{n}+b^{n} d^{n}$ $[$ Using $\left(3\right)]$

$\qquad\qquad \ =c^{n}\left(a^{n}+b^{n}\right)+d^{n}\left(a^{n}+b^{n}\right)$

$\qquad\qquad \ =\left(a^{n}+b^{n}\right)\left(c^{n}+d^{n}\right)$

$\qquad\qquad \ = \text{R.H.S.}$

$\therefore \ \ \left(b^{n}+c^{n}\right)^{2}=\left(a^{n}+b^{n}\right)\left(c^{n}+d^{n}\right)$

Thus, $\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right)$, and $\left(c^{n}+d^{n}\right)$ are in $G.P.$

Answer :

It is given that $a$ and $b$ are the roots of $x^2-3 x+p=0$

$\therefore \ \ a+b=3$ and $a b=p\qquad \ldots \left(1\right)$

Also, $c$ and $d$ are the roots of $x^{2}-12 x+q=0$

$\therefore \ \ c+d=12$ and $c d=q\qquad \ldots\left(2\right)$

It is given that $a, b, c, d$ are in $G.P.$

Let $a=x, b=x r, c=x r^{2}, d=x r^{3}$

From $\left(1\right)$ and $\left(2\right),$ we obtain

$x+x r=3$

$\Rightarrow x\left(1+r\right)=3$

$x r^{2}+x r^{3}=12$

$\Rightarrow x r^{2}\left(1+r\right)=12$

On dividing, we obtain

$\dfrac{x r^{2}\left(1+r\right)}{x\left(1+r\right)}=\dfrac{12}{3}$

$\Rightarrow r^{2}=4$

$\Rightarrow r= \pm 2$

When $r=2, x=\dfrac{3}{1+2}=\dfrac{3}{3}=1$

When $r=-2, x=\dfrac{3}{1-2}=\dfrac{3}{-1}=-3$

Case I:

When $r=2$ and $x=1$,

$a b=x^{2} r=2$

$c d=x^{2} r^{5}=32$

$\therefore \ \ \dfrac{q+p}{q-p}=\dfrac{32+2}{32-2}=\dfrac{34}{30}=\dfrac{17}{15}$

i.e., $\left(q+p\right):\left(q-p\right)=17: 15$

Case II:

When $r=-2, x=1,$

$a b=x^{2} r=-18$

$c d=x^{2} r^{5}=- 288$

$\therefore \ \ \dfrac{q+p}{q-p}=\dfrac{-288-18}{-288+18}=\dfrac{-306}{-270}=\dfrac{17}{15}$

i.e., $\left(q+p\right):\left(q-p\right)=17: 15$

Thus, in both the cases, we obtain $(q+p):(q - p)=17: 15$

Answer :

Let the two numbers be $a$ and $b$.

A.M $=\dfrac{a+b}{2}$ and G.M. $=\sqrt{a b}$

According to the given condition,

$\dfrac{a+b}{2 \sqrt{a b}}=\dfrac{m}{n}$

$\Rightarrow \dfrac{\left(a+b\right)^{2}}{4\left(a b\right)}=\dfrac{m^{2}}{n^{2}}$

$\Rightarrow\left(a+b\right)^{2}=\dfrac{4 a b m^{2}}{n^{2}}$

$\Rightarrow\left(a+b\right)=\dfrac{2 \sqrt{a b} m}{n}\qquad \ldots\left(1\right)$

Using this in the identity $\left(a \text{ - } b\right)^{2}=\left(a+b\right)^{2} - 4 a b$, we obtain

$ \left(a-b\right)^{2}=\dfrac{4 a b m^{2}}{n^{2}}-4 a b=\dfrac{4 a b\left(m^{2}-n^{2}\right)}{n^{2}}$

$ \Rightarrow\left(a-b\right)=\dfrac{2 \sqrt{a b} \sqrt{m^{2}-n^{2}}}{n} \qquad \ldots{\left(2\right)}$

Adding $\left(1\right)$ and $\left(2\right),$ we obtain

$ \begin{aligned} & 2 a=\dfrac{2 \sqrt{a b}}{n}\left(m+\sqrt{m^{2}-n^{2}}\right) \\ \\ & \Rightarrow a=\dfrac{\sqrt{a b}}{n}\left(m+\sqrt{m^{2}-n^{2}}\right) \end{aligned} $

Substituting the value of $a$ in $\left(1\right),$ we obtain

$ \begin{aligned} & b=\dfrac{2 \sqrt{a b}}{n} m-\dfrac{\sqrt{a b}}{n}\left(m+\sqrt{m^{2}-n^{2}}\right) \\ \\ & \ =\dfrac{\sqrt{a b}}{n} m-\dfrac{\sqrt{a b}}{n} \sqrt{m^{2}-n^{2}} \\ \\ & \ =\dfrac{\sqrt{a b}}{n}\left(m-\sqrt{m^{2}-n^{2}}\right) \\ \\ & \therefore \ \ a: b=\dfrac{a}{b}=\dfrac{\dfrac{\sqrt{a b}}{n}\left(m+\sqrt{m^{2}-n^{2}}\right)}{\dfrac{\sqrt{a b}}{n}\left(m-\sqrt{m^{2}-n^{2}}\right)}=\dfrac{\left(m+\sqrt{m^{2}-n^{2}}\right)}{\left(m-\sqrt{m^{2}-n^{2}}\right)} \end{aligned} $

Thus, $a: b=\left(m+\sqrt{m^{2}-n^{2}}\right):\left(m-\sqrt{m^{2}-n^{2}}\right)$

Answer :

$\left(i\right) \ \ $ $5+55+555+\ldots$

Let $S_n=5+55+555+\ldots$. to $n$ terms

$ \begin{aligned} &\qquad =\dfrac{5}{9}\bigg[9+99+999+\ldots \text{ to } n \text{ terms }\bigg] \\ \\ &\qquad =\dfrac{5}{9}\bigg[\left(10-1\right)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots \text{ to } n \text{ terms }\bigg] \\ \\ &\qquad =\dfrac{5}{9}\bigg[\left(10+10^{2}+10^{3}+\ldots \text{ n terms }\right)-\left(1+1+\ldots n \text{ terms }\right)\bigg] \\ \\ &\qquad =\dfrac{5}{9}\bigg[\dfrac{10\left(10^{n}-1\right)}{10-1}-n\bigg] \\ \\ &\qquad =\dfrac{5}{9}\bigg[\dfrac{10\left(10^{n}-1\right)}{9}-n\bigg] \\ \\ &\qquad =\dfrac{50}{81}\left(10^{n}-1\right)-\dfrac{5 n}{9} \end{aligned} $

$\left(ii\right) \ \ $ $0.6+0.66+0.666+\ldots$

Let $S_n=0.6+0.66+0.666+\ldots$ to $n$ terms

$\quad\qquad=6\bigg[0.1+0.11+0.111+\ldots$ to $n$ terms $\bigg]$

$\quad\qquad=\dfrac{6}{9}\bigg[0.9+0.99+0.999+\ldots$ to $n$ terms $\bigg]$

$\quad\qquad=\dfrac{6}{9}\bigg[\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{10^{2}}\right)+\left(1-\dfrac{1}{10^{3}}\right)+\ldots.$ to n terms $\bigg]$

$\quad\qquad=\dfrac{2}{3}\bigg[\left(1+1+\ldots n. \ \text{terms}\right)-\dfrac{1}{10}\left(1+\dfrac{1}{10}+\dfrac{1}{10^{2}}+\ldots n. \ \text{terms}\right)\bigg]$

$\quad\qquad=\dfrac{2}{3}\left[n-\dfrac{1}{10}\left(\dfrac{1-\left(\dfrac{1}{10}\right)^{n}}{1-\dfrac{1}{10}}\right)\right]$

$\quad\qquad=\dfrac{2}{3} n-\dfrac{2}{30} \times \dfrac{10}{9}\left(1-10^{-n}\right)$

$\quad\qquad=\dfrac{2}{3} n-\dfrac{2}{27}\left(1-10^{-n}\right)$

Answer :

The given series is $2 \times 4+4 \times 6+6 \times 8+\ldots n$ terms

$\therefore \ \ n^{\text{th }}$ term $=a_n=2 n \times\left(2 n+2\right)=4 n^{2}+4 n$

$\qquad\quad \ a _ {20}=4\left(20\right)^{2}+4\left(20\right)=4\left(400\right)+80=1600+80=1680$

Thus, the $20^{\text{th }}$ term of the series is 1680 .

Answer :

It is given that the farmer pays Rs 6000 in cash.

Therefore, unpaid amount = Rs 12000 - Rs $6000=$ Rs 6000

According to the given condition, the interest paid annually is $12 \%$ of $6000,12 \%$ of $5500,12 \%$ of $5000, \ldots, 12 \%$ of 500

Thus, total interest to be paid $=12 \%$ of $6000+12 \%$ of $5500+12 \%$ of $5000+\ldots+12 \%$ of 500

$\hspace{3.8cm}=12 \%$ of $\left(6000+5500+5000+\ldots+500\right)$

$\hspace{3.8cm}=12 \%$ of $\left(500+1000+1500+\ldots+6000\right)$

Now, the series $500,1000,1500 \ldots 6000$ is an $A.P.$ with both the first term and common difference equal to $500.$

Let the number of terms of the $A.P.$ be $n$.

$\therefore \ \ 6000=500+\left(n - 1\right) 500$

$\Rightarrow 1+\left(n- 1\right)=12$

$\Rightarrow n=12$

$\therefore \ \ $ Sum of the $A.P=\dfrac{12}{2}\bigg[2\left(500\right)+\left(12-1\right)\left(500\right)\bigg]=6\big[1000+5500\big]=6\left(6500\right)=39000$

Thus, total interest to be paid $=12 \%$ of $\left(500+1000+1500+\ldots+6000\right)=12 \%$ of $39000=$ Rs $4680$

Thus, cost of tractor $=( $ Rs $12000+$ Rs $4680) $ $=$ Rs $16680$

Answer :

Given Shamshad Ali buys a scooter for Rs. 22000 and pay Rs 4000 in cash

$\Rightarrow$ Unpaid amount $=22000-4000=$ Rs $18000 .$

At the end of first year, interest paid $=10 \%$ of $18000$

Amount paid as instalment at the end of $1$ year $=$ Rs $1000.$

So, loan amount left $=$ Rs $18000-1000=$ Rs $17000$

Now, at the end of second year, interest paid $10 \%$ of $17000$

Amount paid as instalment at the end of second year $=$ Rs $1000$

So, loan amount left $=$ Rs $17000-1000=$ Rs $16000$

So, total interest to be paid $=10 \%$ of $18000+10 \%$ of $17000+10 \%$ of $16000+\ldots+10 \% \text { of } 1000 . $

$\hspace{3.7cm} =10 \% \text { of }(18000+17000+16000+\ldots 1000)$

$\hspace{3.7cm}=10 \% \text { of }(1000+2000+3000+\ldots+18000) \qquad\ldots .(1)$

Here, $1000,2000,3000, \ldots 18000$ forms an $A.P.$ with both the first term and common difference equals to $1000 .$

Let, the number of terms of $A.P.$ be $n$.

$\therefore 18000=1000+(\mathrm{n}-1)(1000) $

$\Rightarrow \mathrm{n}=18 $

$\therefore \mathrm{S}_ {18}=1000+2000+\ldots+18000 $

$\qquad\quad=\dfrac{18}{2}\big[2(1000)+(18-1)(1000)\big]$

$\qquad\quad=9\big[2000+17000\big]=171000$

So, by eqn(1),

$\text { Total interest paid }=10 \% \text { of }(18000+17000+16000+\ldots+1000) $

$\hspace{3.1cm}=10 \% \text { of Rs. } 1,71,000$

$\hspace{3.1cm}=\text { Rs. } 17,100$

So, the cost of scooter to him is Rs. $22000+17100=39100$

Answer : According to question, one person writes letter to his $4$ friends Now, each of these $4 $friends will write letter to their $4$ friends. So, the number of letters written here will be $4{ }^2$ This continues….

So, the number of letters written forms a G.P. $4,4^2, 4^3, \ldots .4^8$

Here, $a=4, r=4, n=8$

We know Sum of $n$ terms of G.P. $=\dfrac{a\left(r^n-1\right)}{r-1}$

So, sum of 8 terms of G.P. is $\dfrac{4\left(4^8-1\right)}{4^{-1}}=\dfrac{4(65536-1)}{3}=\dfrac{4 \times 65535}{3}=87380$

So, the number of letter written in $8$ th set of letter is $87380.$

Cost to mail one letter is $50$ paise

So, cost to mail $87380$ is Rs $\dfrac{50}{100} \times 8738 0=$ Rs. $43690$

Answer :

It is given that the man deposited Rs $10000$ in a bank at the rate of 5\% simple interest annually.

$\therefore \ \ $ Interest in first year $ =\dfrac{5}{100} \times Rs 10000=Rs 500 $

$\therefore \ \ $ Amount in $15^{\text{th }}$ year $=Rs$ $ 10000+\underbrace{500+500+\ldots+500} _ {14 \text{ times }} $

$\hspace{3.3cm}=$ Rs $10000+14 \times$ Rs $500$

$\hspace{3.3cm}=$ Rs $10000+$ Rs $7000$

$\hspace{3.3cm}=$ Rs $17000$

Amount after $20$ years $=$ Rs $10000+\underbrace{500+500+\ldots+500} _ {20 \text{ times }} $

$\hspace{3cm}=$ Rs $10000+20 \times$ Rs $500$

$\hspace{3cm}=$ Rs $10000+$ Rs $10000$

$\hspace{3cm}=$ Rs $20000$

Answer :

Cost of machine $=$ Rs $15625$

Machine depreciates by $20\%$ every year.

Therefore, its value after every year is $80 \%$ of the original cost i.e., $\dfrac{4}{5}$ of the original cost.

$\therefore \ \ $ Value at the end of 5 years $=15625 \times \underbrace{\dfrac{4}{5} \times \dfrac{4}{5} \times \ldots \times \dfrac{4}{5}} _ {5 \text{ times }}=5 \times 1024=5120$

Thus, the value of the machine at the end of 5 years is Rs $5120 .$

Answer :

Let $x$ be the number of days in which $150$ workers finish the work.

According to the given information,

$150 x=150+146+142+\ldots .\left(x+8\right)$ terms

The series $150+146+142+\ldots .\left(x+8\right)$ terms is an $A.P.$ with first term $146 ,$ common difference $- 4$ and number of terms as $\left(x+8\right)$

$ \begin{aligned} & \Rightarrow 150 x=\dfrac{\left(x+8\right)}{2}[2\left(150\right)+\left(x+8-1\right)\left(-4\right)] \\ \\ & \Rightarrow 150 x=\left(x+8\right)[150+\left(x+7\right)\left(-2\right)] \\ \\ & \Rightarrow 150 x=\left(x+8\right)\left(150-2 x-14\right) \\ \\ & \Rightarrow 150 x=\left(x+8\right)\left(136-2 x\right) \\ \\ & \Rightarrow 75 x=\left(x+8\right)\left(68-x\right) \\ \\ & \Rightarrow 75 x=68 x-x^{2}+544-8 x \\ \\ & \Rightarrow x^{2}+75 x-60 x-544=0 \\ \\ & \Rightarrow x^{2}+15 x-544=0 \\ \\ & \Rightarrow x^{2}+32 x-17 x-544=0 \\ \\ & \Rightarrow x\left(x+32\right)-17\left(x+32\right)=0 \\ \\ & \Rightarrow\left(x-17\right)\left(x+32\right)=0 \\ \\ & \Rightarrow x=17 \text{ or } x=-32 \end{aligned} $

However, $x$ cannot be negative.

$\therefore \ \ x=17$

Therefore, originally, the number of days in which the work was completed is $17.$

Thus, required number of days $=\left(17+8\right)=25$