Here \(\mathbf{t}_{1} =\frac{2 m+1}{m} \mathrm{t}_{2}=\frac{2 m-1}{m} \)
\(\mathrm{~d} =\mathrm{t}_{2}-\mathrm{t}_{1} \)
\(=\frac{2 m-1}{m}-\frac{2 m+1}{m} \)
\(=\frac{2 m-1-2 m-1}{m} \)
\(=\frac{-2}{m} \)
\(\text {Now } t_{n} =a+(n-1) d \)
\(t_{n} =\left[\frac{2 m+1}{m}\right]+(n-1)\left[\frac{-2}{m}\right] \)
\(=\left[\frac{2 m+1}{m}\right]+\left[\frac{-2 n}{m}\right]-1\left[\frac{-2}{m}\right] \)
\(=\frac{2 m+1}{m}-\frac{2 n}{m}+\frac{2}{m} \)
\(=\frac{2 m+1-2 n+2}{m} \)
\(\text {Thus } \mathrm{n}^{\text {th }} \text { term }=\frac{2 m-2 n+3}{m}\)
\(\therefore 6^{\text {th }} \text { term } t_{6}=\frac{2 m-2(6)+3}{m}\)
\(6^{\text {th }} \text { term }=\frac{2 m-9}{m}\)

\(\text { Given } \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \text { are in A.P. }\)
\(\therefore \mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2}\)
\(\frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}\)
\(\frac{(b+c)-(c+a)}{(c+a)(b+c)}=\frac{(c+a)-(a+b)}{(a+b)(c+a)}\)
\(\frac{b+c-c-a}{c b+a b+c^{2}+a c}=\frac{c+a-a-b}{c a+b c+a^{2}+a b} \)
\(\frac{b-a}{a b+c b+a c+c^{2}}=\frac{c-b}{a b+b c+c a+a^{2}} \)
\((b-a)\left(a b+b c+c a+a^{2}\right) \)
\(=(c-b)\left(a b+c b+a c+c^{2}\right) \)
\(a b^{2}+b^{2} c+c a b+a^{2} b-a^{2} b-a b c-a^{2} c-a^{3} \)
\(=a b c+c^{2} b+a c^{2}+c^{3}-a b^{2}- \)
\(c b^{2}-a b c-b c^{2} \)
\(a b^{2}+b^{2} c-a^{2} c-a^{3}=a c^{2}+c^{3}-a b^{2}-c b^{2} \)
\(b^{2} c-a^{2} c+a b^{2}-a^{3}=c^{3}-c b^{2}+a c^{2}-a b^{2} \)
\(c\left(b^{2}-a^{2}\right)+a\left(b^{2}-a^{2}\right) \)
\(=c\left(c^{2}-b^{2}\right)+a\left(c^{2}-b^{2}\right) \)
\(\left(b^{2}-a^{2}\right)(c+a) =\left(c^{2}-b^{2}\right)(c+a) \)
\(b^{2}-a^{2} =c^{2}-b^{2} \)
\(\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2} \text { of the sequence } \mathrm{a}^{2}, \mathrm{~b}^{2}, \mathrm{c}^{2}\therefore \mathrm{a}^{2}, \mathrm{~b}^{2}, \mathrm{c}^{2} \text { are in A.P. }\)

Given a₅ + a₉ = 72 and a₇ + a₁₂ = 97
a + 4d + a + 8d = 72 and a + 6d + a + 11d = 97
2a + 12d = 72..(1) and 2a + 17 d, = 97
Solving (1) and (2)
2a + 12 d = 72
2a + 17d = 97
\((2)-(1) \Rightarrow 5 d=25\)
\(\mathrm{d} =5 \)
\(2 \mathrm{a}+12 \mathrm{~d} =72 \)
\(2 \mathrm{a}+12(5) =72 \)
\(2 \mathrm{a}+60 =72 \)
\(2 \mathrm{a} =72-60=12 \)
\(\mathrm{a} =\frac{12}{2}=6 \)
The A.P. is a, a + d, a + 2d.a + 3d...
6, 6 + 5, 6 + 2(5),6 + 3(5) ...
6, 11, 16,21,...

\(t_{n}=a+(n-1) d\)
10^th term of an A.P. is 52.
\(t_{10}=a+(10-1) d=52\)
\(a+9 d=52\)
16^th term is 82
a + 15d = 82
a + 9d = 52
Solving (1) and (2)
(2) - (1), 6d = 30
d = 5
a + 9d = 52 from (l)
a +9(5) = 52
a + 45 = 52
a - 52 - 45 = 7
t₃₂ = a + 31d
= 7 + 31(5) = 7 + 155
t₃₂ = 162
The general term a_n = 5n + 2

Given  \(a_{n}=n^{3}-6 n^{2}+11 n-6\)
\(\text { Put } \mathrm{n}=1, \mathrm{a}_{1} =1^{3}-6(1)^{2}+11(1)-6 \)
\(=1-6+11-6=12-12 \)
\(\mathrm{a}_{1} =0 \)
\(\text { Put } \mathrm{n}=2, \mathrm{a}_{2} =2^{3}-6(2)^{2}+11(2)-6 \)
\(=8-(6 \times 4)+22-6 \)
\(=8-24+22-6=30-30 \)
\(\mathrm{a}_{2} =0 \)
\(\text { Put } \mathrm{n}=3, \mathrm{a}_{3} =3^{3}-6(3)^{2}+11(3)-6 \)
\(=27-54+33-6=60-60 \)
\(\mathrm{a}_{3} =0 \)
So we have a₁ = a₂ = a₃ = 0
That is the value of the cubic polynomial \(a_{n}=n^{3}-6 n^{2}+11 n-6\)
becomes zero for n = 1, 2, 3
\(\therefore(n-1),(n-2) \text { and }(n-3) \text { are factors of } a_{n}\)
\(\therefore a=(n-1)(n-2)(n-3) \text { for which } a_{n}>0 \text { for }\text { all } n>3\)
Other terms are Positive

Given \(a_{1}=4\)
\(a_{n}=4 a_{n-1}+3 \text { for } n>1\)
\(\text { Put } n=2 ; a_{2}=4 a_{2-1}+3=4 a_{1}+3\)
\(=4(4)+3=16+3\)
\(a_{2}=19\)
\(\text { Put } n=3 ; \quad a_{3}=4 a_{3-1}+3=4 a_{2}+3\)
\(=4(19)+3=76+3\)
\(a_{3}=79\)
\(\text { Put } n=4 ; a_{4}=4 a_{4-1}+3=4 a_{3}+3\)
\(=4(79)+3=316+3\)
\(a_{4}=319\)
\(\text { Put } n=5 ; a_{5}=4 a_{5-1}+3=4 a_{4}+3\)
\(=4(319)+3=1276+3\)
\(a_{5}=1279\)
\(\text { Put } n=6 ; a_{6}=4 a_{6-1}+3=4 a_{5}+3\)
\(=4(1279)+3=5116+3\)
\(a_{6}=5119 \)
\(\therefore \text { The next five terms are } a_{n}=19, a_{3}=79,a_{4}=319, a_{5}=1279, a_{6}=5119\)

9x \(\equiv\) 5 (mod13)
9x - 5 = 13 k for some integer k
\(x=\frac{13 k+5}{9} \text { for some integer } k\)
When we put 1, 10, 19,.... the 13k + 5 is divisible by 13.
\(x=\frac{13(1)+5}{9}=2 \)
\(x=\frac{13(10)+5}{9}=15 \)
\(x=\frac{13(19)+5}{9}=28 \)
\(x=\frac{13(28)+5}{9}=41 \)
The solutions are 2, 15, 28, 41,....

L.C.M. = 12 x 10 = 120
L.C.M. of 2,4,6,8,10, 12 is 120
\(\therefore\) The bells will toll together after every 120 sec. i.e., 2 min.
In 30 minutes they will toll together \(\left(\frac{30}{2}+1\right)=16\) times

Smallest number of five digits is 10000.
Required number must be divisible by 16,24,35 and 54

\(\text { L.C.M. }=2^{4} \times 3^{3}\)
= 16 x 27 = 432
L.C.M. of 16, 24,36,54 = 432
\(\therefore\) Required number is divisible by 432.

On dividing 10000 by 432,we get remainder = 64
Required number = 10000 + (432 - 64)
= 10358

Required number = H.C.F. of(1356 - 12),(1868 - 12), (2764 - 12)
= H.C.F. of 1344,1856, 2752

\(2752=2^{6} \times 43^{1}\)
\(1344=2^{6} \times 3^{1} \times 7^{1} \)
\(1856=2^{6} \times 29^{1} \)
\(\text { H.C.F. is } 2^{6}=64\)
\(\therefore \text { The required number }=64\)

The number of participants in each room is the H.C.F. of 60, 84 and 108.
First we find the H.C.E of 60 and 84, by applying
Euclid's Division Algorithm.
84 = 60 x 1 +24
60 = 24 x 2+ 12
24 = 12 x 2 + 0
The remainder = 0
\(\therefore\) H.C.F. of 60 and 84 is 12.
Now applying Euclid's Division Algorithm to 12 and 108
108 = 12 x 9 + 0
Remainder = 0
\(\therefore\) H.C.F. of 12 and 108 = 12
\(\therefore\) H.C.F. (60,84, 108) = 12.
\(\therefore\) In each room, the minimum 12 number of participants can be seated.
Total number of participants = 60 + 84 + 108
= 252
Number of rooms required \(=\frac{252}{12}=21\)

Applying Euclid's division Algorithm to find the H.C.E of 56 and 72, we have
72 = 56 x 1 + 16 ...(1)
56 = 15 x 3 + 8...(2)
16 = 8 x 2 + 0
Remainder = 0
H.C.F. of 56 and 72 = 8
From (2) we have
8 = 56 - 16 x 3
8 = 56 - (72 -56 x 1) x 3
[\(\because\) from(1) 16 = 72 - 56 x 1]
= 56 - (3 x 72) + 3 x 56
= 56(1 +3)-(3 x 72)
= 56(4) + 72(-3)
Comparing with d = 55x + 72y,we have
x = 4 and y = -3

Let us find the H.C.F. of 210 and 55, Applying
Euclid's Division Algorithm
210 - 55 x 3 + 45
55 = 45 x 1 + 10
45 = 10 x 4 + 5
10 = 5 x 2 + 0
Remainder = 0
H.C.F. of 210 and 55 is 5
Given 5 - 210 x 5 + 55y
5 - 210 x 5 = 55y
55y = 5 - 1050 = -1045
\(y=\frac{-1045}{55}\)
\(y = -19\)

Let m and n be any integers, then
x = 2m + 1 and Y = 2n + 1 since x and y are odd
\(x^{2}+y^{2} =(2 m+1)^{2}+(2 n+1)^{2} \)
\(=4 m^{2}+4 m+1+4 n^{2}+4 n+1 \)
\(=4\left(m^{2}+n^{2}\right)+4(m+n)+(1+1) \)
\(=4\left(m^{2}+n^{2}\right)+4(m+n)+2 \)
\(=4 q+2, \text { where } q=\left(m^{2}+n^{2}\right)+(m+n) \)
\(=4 q+2 \)
\(x^{2}+y^{2}=2[2 a+1] \text { which is an even number. }\)
Thus 4q + 2 is an even number, which is not divisible by 4.
i.e., It leaves the remainder 2.
Hence x² + y² is even but not divisible by 4.

First we will Prove "any odd positive integer n is of the form 4q + 1 or 4q + 3, where q is some integer"'
By Euclid's division Lemma, 
If 'a' and 'b' are two positive integers then
a = bq + r where \(0 \leq \mathrm{r}<|\mathrm{b}|\)
Suppose the positive integer be 'a' and b = 4
then a = 4q + r where \(0 \leq \mathrm{r}<|\mathrm{4}|\)
\(a=40 . a=4 q+1 ; a=4 q+2 ; \quad a=4 q+3\)
\(=2(2 q) ; a=4 q+1=2(2 q+1) ; =\text { odd }\)
\(=\text { even; } =\text { odd } ; =\text { even }\)
Any positive odd integer is of the form 4q + 1 or 4q + 3
Case (I)
\(\text { If } a =4 q+1 \)
\(a^{2} =(4 q+1)^{2} \)
\(=16 q^{2}+8 q+1 \)
\(=8 q(2 q+1)+1 \)
\(=8 m+1 \text { where } m=q(2 q+1) \)
Case (II)
\(\text { If } a=4 q+3\)
then \(a^{2} =(4 q+3)^{2}=16 q^{2}+24 q+9 \)
\(=8\left[2 q^{2}+3 q\right]+8+1 \)
\(=8\left[2 q^{2}+3 q+1\right]+1 \)
\(=8 m+1 \text { where } m=2 q^{2}+3 q+1 \)
We conclude that the square of an odd positive integer is of the form 8q + 1, for same integer q.