Given point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).
So,   AP = BP
\(\Rightarrow\)   AP² = BP²
\(\Rightarrow\) (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
       \(\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]\)
\(\Rightarrow\)   x² + 49 - 14x + y² + 1 - 2y
                = x² + 9 - 6x + y² + 25 - 10y
\(\Rightarrow\)  -14x - 2y + 50 = -6x - 10y + 34
\(\Rightarrow\) -6x - 10y + 14x + 2y = 50 - 34
\(\Rightarrow\)              8x - 8y = 16
\(\Rightarrow\)                 x - y = 2
                                [dividing by 8 on both sides]
Hence, the relation between x and y is x - y = 2.

Let S_n = 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.
Therefore,   \(\mathrm{S}_{n}=\frac{n(1+n)}{2} \text { or } \mathrm{S}_{n}=\frac{n(n+1)}{2}\)
So, the sum of first n positive integers is given by
\(\mathrm{S}_{n}=\frac{n(n+1)}{2}\) 

Using formula for area of a triangle, we get 
Area of triangle
= \(\frac{1}{2}\)|a{c+a)-(a+b)}+b{(a+b)-(b+c)}+c{(b+c)-(c+a)}|
= \(\frac{1}{2}\)|a{(c+a)-(a+b)}+b{(a+b)-(b+c)}+c{(b+c)-(c+a)}|
= \(\frac{1}{2}\)|a{c+a-a-b}+b{a+b-b-c}+c{b+c-c-a}|
= \(\frac{1}{2}\)|a(c-b)+b(a-c)+c(b-a)|
= \(\frac{1}{2}\)|ac-ab+ba-bc+cb-ca|=0

Yes

The odd numbers between 0 and 50 are 1,3, 5,., 49 which form an AP.
Here, first term (a) = 1, last term (l) = 49
and common difference (d) = 3 - 1 = 2.
Let there be n numbers in the AP.
Then, nth term (a_n) = a + (n - 1) d = l
\(\Rightarrow\) 1 + (n - 1) (2) = 49
\(\Rightarrow\) n - 1 = 24 \(\Rightarrow\) n = 25
Now, sum of n terms, \(S_n=\frac{n}{2}(a+l)\)
\(\therefore\) Sum of 25 terms, \(\begin{aligned} S_{25} & =\frac{25}{2}(1+49)=\frac{25}{2} \times 50 \\ \end{aligned}\)
\(\begin{aligned} & =25 \times 25=625 \end{aligned}\)

(-3, 1)

Yes, 6

Area of quadrilateral ABCD is

= \(\frac{1}{2}\)|{(-5)(-5)+(-4)(-6)+(-1)(5)+4(7)}-{(-4)(7)+(-1)(-5)+4(-6)+(-5)(5)}|
= \(\frac{1}{2}\)|(25+24-5+28)-(-28+5-24-25)|
=\(\frac{1}{2}\)|72-(-72)|
=\(\frac{1}{2}\)|144|
=\(\frac{1}{2}\)×144
=72 sq.units

Here, S₁₄ = 1050, n = 14, a = 10.
As  \(\begin{aligned} \mathrm{S}_n & =\frac{n}{2}[2 a+(n-1) d] \\ \end{aligned}\)
So, \(\begin{aligned} 1050 & =\frac{14}{2}[20+13 d]=140+91 d \end{aligned}\)
i.e., 910 = 91d
or, d = 10
Therefore, a₂₀ = 10 + (20 - 1) \(\times\) 10 = 200, i.e. 20th term is 200.

The list of two-digit numbers divisible by 3 is :
12, 15, 18, . . . , 99
Is this an AP? Yes it is. Here, a = 12, d = 3, a_n = 99.
As   a_n = a + (n – 1) d,
we have   99 = 12 + (n – 1) x 3
i.e., 87 = (n – 1) x 3
i.e.,  \(n-1=\frac{87}{3}=29\)
i.e., n = 29 + 1 = 30
So, there are 30 two-digit numbers divisible by 3.

It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a₂ − a₁ = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
S_n = 200
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(200=\frac{n}{2}[2(20)+(n-1)(-1)]\)
400 = n (40 − n + 1)
400 = n (41 − n)
400 = 41n − n²
n² − 41n + 400 = 0
n² − 16n − 25n + 400 = 0
n (n − 16) −25 (n − 16) = 0
(n − 16) (n − 25) = 0
Either (n − 16) = 0 or n − 25 = 0
n = 16 or n = 25
a_n = a + (n − 1)d
a₁₆ = 20 + (16 − 1) (−1)
a₁₆ = 20 − 15
a₁₆ = 5
Similarly,
a₂₅ = 20 + (25 − 1) (−1)
a₂₅ = 20 − 24
= −4
Clearly, the number of logs in 16^th row is 5. However, the number of logs in 25^th row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

a = 8, d = 1/3 years, S_n = 168
\( { S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d]\)
\(\Rightarrow \quad 168=\frac { n }{ 2 } \left[ 2(8)+(n-1)\frac { 1 }{ 3 } \right] \)
n² + 47n - 1008 = 0
\(\Rightarrow\) n² + 63n - 16n - 1008= 0
\(\Rightarrow\) (n - 16)(n + 63) = 0
\(\Rightarrow\) n = 16 or n = - 63
n = 16
(n cannot be negative)
Age of the oldest participant = a + 15d = 13 years

AC : CD : DB = 1 : 1 : 1.Hence, AD : DB = 2 : 1
First, find D using section formula and then find point. C using mid-point formula between A and D, as C is their mid-point.
\(\left( -1,\frac { 1 }{ 2 } \right) ,\left( 0,3 \right) ,\left( 1,\frac { 11 }{ 2 } \right) \)

We have, 3^rd term = 4 and 9^th term = -8 i.e., a + 2d = 4 ........(i)
and a + 8d = -8 .........(ii)
Solving (1) and (2), we get
d = -2, a = 8
(i) (c)
(ii) (d)
(iii) (b): Let t_n = -160 \(\Rightarrow\) a + (n - 1) d = -160
\(\Rightarrow\) 8 + (n - 1)(-2) = -160 \(\Rightarrow\) (n - 1)(-2) = -168
\(\Rightarrow\) n - 1 = 84 \(\Rightarrow\) n = 85
So, t₈₅ = -160
(iv) (a)
(v) (a): t₇₅ = a + 74d = 8 + 74( -2) = -140