We know that, a rational number has either terminating or non-terminating repeating decimal expansion.
Here, 43.123456789 has terminating decimal expansion. So, it represents a rational number.
i.e. 43123456789 =\(\frac{43123456789}{1000000000}\)= \(\frac{p }{q}\)
Thus, q = 109, whose factors are 2⁹ x 5⁹.

Sum of zeroes \(=-\frac { Coefficient\ of \ x }{ Coefficient \ of \ { x }^{ 2 } } \) 
\(\therefore \quad a + b = - a\)
\(\Rightarrow \quad 2a + b = 0\)
Product of zeroes = \(​​=-\frac { Constant \ term }{ Coefficient \ of \ { x }^{ 2 } } \)
\(\therefore \quad ab = b\)
\(\Rightarrow \quad a=1\)
then b = - 2

a_n = 1 - 4n
\(\Rightarrow\)  a₁ = 1 - 4 x 1 = -3
a₂ = 1 - 4 x 2 = -7
d = a₂ - a₁
= - 7 - ( - 3 ) = - 4
a₂₅ = a + 24d
- 3 = 24 x ( - 4 ) = - 99
Now,    S₂₅ = \(\frac{25}{2}({a}_{1}+{a}_{25})\)
\(=\frac{25}{2}(-3-99)\)
= 25 x ( - 51 ) = - 1275

= \(-\sqrt{3},-7 / \sqrt{3}\)

Let n be any +ve integer, then
n = 3q + r, r = 0, 1, 2
n = 3q or 3q + 1 or 3q + 2
Case I:
When, n = 3q,which is divisibleby 3
n + 2 = 3q + 2, which is not divisible by 3
n + 4 = 3q + 4, which is not divisible by 3
Case II:
When, n = 3q + 1, which is not divisible by 3 'h
n + 2 = 3q + 1 + 2 = 3q + 3, which is divisible by 3
n + 4 = 3q + 1 + 4 = 3q + 5,  which is not divisible by 3
Case III:
When, n = 3q + 2, which is not divisible by 3
n + 2 = 3q + 2 + 2 = 3q + 4, which is not divisible by 3
n + 4 = 3q + 2 + 4 = 3q + 6, which is divisible 3
Case I, II and III \(\Rightarrow\) One and only one out of n, n + 2 or n + 4 is divisible by 3.

32

Considers \(d_1=d_2+4\) or \(d_2=d_1+4\), where \(d_1\) and \(d_2\) are the common differences of the two arithmetic progressions.
Also, \(a_{124}=b_{42} \Rightarrow a_1+123 d_1=b_1+41 d_2\) where \(a_1, a_{124}, b_1\) and \(b_{42}\) are the \(1 \mathrm{st}, 124 \mathrm{th}, 1\) st and 42 nd terms of the two arithmetic progressions, respectively.
\(\Rightarrow \quad a_1+123 d_1=a_1+41 d_2 \quad\left[\because a_1=b_1\right]\)
Solve the above two equations and the value of \(d_1=-2\) and \(d_2=-6\) or \(d_1=2\) and \(d_2=6\).
One set of possible value of common difference is as follows
\(a_{124}  =b_{42} \)
\(\Rightarrow a_1+123\left(4+d_2\right)  =b_1+41 d_2\)
\(\Rightarrow \quad 492+123 d_2  =41 d_2 \Rightarrow d_2=-6\)

Given, (a-b),a and (a+b) are the zeroes of the polynomial x³-3x²+x+1.
On comparing given polynomial with
Ax³+Bx²+cx²+D, we get
A=1, B=-3, C=1 and D=1
Now, sum of zeroes=(a-b)+a+(a+b)=-B/A
⇒ \(3a=\frac { -(-3) }{ 1 } =3\)              [∵ B=-3, A=1]
⇒ 3A=3⇒A=1
Sum of product of zeroes taken two at a time
=a(a-b)+a(a+b)+(a+b)(a-b)=\(\frac { C }{ A } \)
⇒ a²-ab+a²+ab+a²-b²=1/1  [∵ C=1, A=1]
⇒ 3a²-b²=1⇒3(1)²-b²=1      [from Eq.(i)]
⇒ 3-b²=1⇒b²=2⇒b=\( \pm \sqrt { 2 } \)
Hence, a=1 and b=\( \pm \sqrt { 2 } \)

Let a be any positive integer, then apply Euclid's division lemma for a and b = 3

Yes, 6

(i) Given, p(t) = 20t - 6t²
For zeroes.
p(t) = 0
\(\Rightarrow\) 20t - 16t² = 0
\(\Rightarrow\)4t(5 - 4t) = 0
\(\Rightarrow\) 4t = 0 or 5 - 4t = 0
\(\Rightarrow\)   t = 0, t = \(\frac{5}{4}\)
\(\therefore\) Zeros of p(t) are 0 and \(\frac{5}{4}\).
(ii) (a) Because 0 and \(\frac{5}{4}\) are zeroes of polynomial and p(t) should increase when height and time are increasing.
(iii) (a) At t = \(\frac{3}{2}\)
\(p\left(\frac{3}{2}\right)=20 \times \frac{3}{2}-16 \times\left(\frac{3}{2}\right)^2=30-36=-6\)
\(\Rightarrow\) Height = 6 cm
'-ve' denote graph is below the X-axis.
(b) \(\therefore\) Distance = Speed \(\times\)Time
\(=20\times \frac{5}{4}\)
= 25 cm

(a) (ii)
96 = 2⁵ x 3
240 = 2⁴ x3  x5
(b) (iii)
Total number of books = 96 +240+336=672
Number of books in each stack = 48
\(\therefore\) Number of stacks formed -= \(\frac{672}{48}=14\) 
(c) (i)
Number of mathmatics books = 336
Number of stacks of mathematics books formed = \(\frac{336}{48}\) 
= 7
(d) (iv)
Number of books in each stack of english books = 48
Thickness of each english book = 3 cm
\(\therefore\)  Height of each stack of english books = (48X3) cm
= 144cm
(e) (iii)
Number of books in a stack of hindi books = 48
Weight of each hindi book = 1.5kg
\(\therefore\) The weight of books in a stack of hindi books
= (48X1.5)kg = 72kg

(a) (i)
The given AP is 6,12,18...,
The common difference = 12-6 =6.
(b) (ii)
In the given AP, we have:a=d=6
\(\therefore a_{n}=a+(n-1) d=6+(n-1) 6=6+6 n-6=6 n\) 
(c) (ii)
The number of trees planted by the students of all the sections of class
= 8th term of the given AP
= 6n = 6X8=48
(d) (ii)  T
otal number of trees planted by class 12 students
= 6 X 12 = 72
(e) (iii)
3rd term from the end = \((n-3+1) \text { th term }\) 
= (12-3+1) th term = 10th term
= 6X10=60