Here a = 24, d = 21-24 = -3, S_n = 78. We need to find n.
We know that,
S_n= \(\frac{n}{2}\) (2a + (n - 1)d)
78 = \(\frac{n}{2}\) 48 + (n -1)(-3))
78 = \(\frac{n}{2}\) 2(51-3n)
or 3n² - 51n + 156 = 0
n² -17n + 52 = 0
(n - 4)(n - 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

The number of rose plants in the 1st, 2nd, 3rd, . . . rows are
23,21, 19, ... 5
It forms an A.P.
Let the number of rows in the flower bed be n.
Then a = 23, d = 21 - 23 = -2/a = 5.
As, a_n = a + (n - 1)d i.e. t_n = a + (n - 1)d
We have 5 = 23 + (n - 1)(-2)
i.e. -18 = (n - 1)(-2)
n = 10
ஃ There are 10 rows in the flower bed.

The natural numbers between 300 and 600 which are divisible by 7 are 301, 308, 315, …, 595.
The sum of all natural numbers between 300 and 600 is 301 + 308 + 315 +...+ 595
The terms of the above series are in A.P.
First term a = 301; common difference d = 7; Last term l = 595.
\(n=\left( \frac { l-a }{ d } \right) +1=\left( \frac { 595-301 }{ 7 } \right) +1=43\)
Since, \({ S }_{ n }=\frac { n }{ 2 } \left[ a+l \right] \), we have \({ s }_{43 }=\frac { 43 }{ 2 } \left[ 301+595 \right] \) = 19264 

Since 70000 is divisible by 7
70000 \(\equiv \) 0 (mod 7)
70000 + 4 \(\equiv \) 0 + 4 (mod 7)
70004 \(\equiv \)  4 (mod 7)
Therefore, the remainder when 70004 is divided 7 is 4 
Since 777 is divisible by 7
777 \(\equiv \) 0 (mod 7)
777 + 1 \(\equiv \) 0 + 1 (mod 7)
778 \(\equiv \) 1 (mod 7)
Therefore, the remainder when 778 is divided by 7 is 1.

To find HCF of three given numbers, first we have to find HCF of the first two numbers.
To find HCF of 396 and 504
Using Euclid’s division algorithm we get 504 = 396  x  1 + 108
The remainder is 108 \(\neq \) 0
Again applying Euclid’s division algorithm 396  = 108  x 3 + 72 
The remainder is 72 \(\neq \) 0
Again applying Euclid’s division algorithm 108  = 72 x 1 + 36 
The remainder is 36 \(\neq \) 0
Again applying Euclid division algorithm 72 = 36 x 2 + 0
Here the remainder is zero. Therefore HCF of 396 , 504 = 36, To find the HCF of 636 and 36
Using Euclid’s division algorithm we get 636 = 36 x 17 + 24
The remainder is 24 \(\neq \) 0
Again applying Euclid's division algorithm 36 = 24 x 1 + 12
The remainder is 12 \(\neq \) 0
Again applying Euclid's division algorithm 24 = 12 x 2 + 0
Here the remainder is zero. Therefore HCF of 636,36 = 12
Therefore Highest Common Factor of 396, 504 and 636 is 12. 

Let us assume the opposite, (1) \(\sqrt { 3 } \) is irrational.
Hence \(\sqrt { 3 } =\frac { p }{ q } \)
Where p and q (q ≠ 0) are co-prime (no common factor other than 1)
Hence, \(\sqrt { 3 } =\frac { p }{ q } \)
\(\sqrt { 3 } \)q = p
Squaring both side
\({ (\sqrt { 3 }q ) }^{ 2 }={ p }^{ 2 }\)
3q² = p²
\({ q }^{ 2 }=\frac { p }{ 3 } \)
Hence, 3 divides p² So 3 divides p also   .....(1)
Hence we can say
\(\frac{p}{3}\) = c where c is some integer
s, p =p²
Putting p = 3c
3q² = (3c)²
3q² = 9c²
q2 = \(\frac13\) x 9c²
q² = 3c²
\(\frac{9^2}{3}\) = c²
Hence 3 divides q²
So, 3 divides q also  ...(2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 } \) is irrational.

Since 12576 > 4052 we apply the division lemma to 12576 and 4052, to get
12576 = 4052 x 3 + 420.
Since the remainder 420 ≠ 0, we apply the division lemma to 4052
4052 = 420 x 9 + 272.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get
420 = 272 x 1 + 148, 148 ≠ 0
∴ Again by division lemma
272 = 148 x 1 + 124, here 124 ≠ 0
∴ Again by division lemma
148 = 124 x 1 + 24, Here 24 ≠ 0
∴ Again by division lemma
124 = 24 x 5 + 4, Here 4 ≠ 0
∴ Again by division lemma
24 = 4 x 6 + 0.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.
∴ The HCF of 12576 and 4052 is 4.

15 \(\equiv \) 3 (mod d) means 15 - 3 = kd, for some integer k,
12 = kd
gives d divides 12.
The divisors of 12 are 1,2,3,4,6,12. But d should be larger than 3 and so the possible values for d are 4, 6, 12.

Yes, the given number is a composite number, because
7 x 5 x 3 x 2 + 3 = 3 (7 x 5 x 2 + 1) = 3 x 71
Since the given number can be factorized in terms of two primes, it is a composite number.

Since 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
2 is a factor of 6ⁿ. So, 6 n is always even.
But any number whose last digit is 5 is always odd.
Hence, 6ⁿ cannot end with the digit 5 

All the integers 'a' must be either even or odd.
If it is even then a = 2q.
If it is odd then a = 2q + 1
Case 1:
lf a = 2q
a² = (2q)²
a² = 4q²,  remainder 0 when divided by 4.
Case 2:
If a = 2q+ 1
a² = (2q + 2)²
= 4q² + 4q + 1
= 4q ( q + 1) + 1
a² = 4m + 1 Where m = q (q + 1) is an integer
It is of the form bq + 1 where 1 is the remainder when divided by 4.
The square of any integer leaves the remainder either 0 or 1 when divided by 4.