Here the value of n is not given. But the last term is given. From this, we can find the value of n.
Given a = 0.40 and l = 1, we find d = 0.43 - 0.40 = 0.03
Therefore, n = \(\left( \frac { l-a }{ d } \right) +1\)
= \(\left( \frac { 1-0.40 }{ 0.03 } \right) +1=21\)
Sum of first n terms of an A.P S_n = \(\frac { n }{ 2 } \left[ a+l \right] \)
Here, n = 21. Therefore, S₂₁ = \(\frac { 21 }{ 2 } \left[ 0.40+1 \right] =14.7\)
So, the sum of 21 term of the given series is 14,7.

Let the three consecutive terms be a - d, a, a + d
Given their sum is 27
(a - d) + a + (a + d) = 27
a - d + a + a + d = 27
3a = 27
\(a=\frac{27}{3}=9\)
Product = 288
(a - d) a (a + d) = 288
a(a² - d²) = 288
9(9² - d²) = 288
\(81-d^{2}=\frac{288}{9}\)
81 - d² = 32
d² = 81 - 32 = 49
d x d = 7 x 7
d = \(\pm\)7
(i) a = 9, d = 7, The three terms are,
= 9 -7, 9, 9 + 7
= 2,9, 16
(ii) a = 9, d = -7, The three terms are
9-(-7), 9, 9-7 \(\Rightarrow\) 16,9,2
The required three consecutive terms of the A.P are 2,9,16.

Let the amount received by the three children be in the form of A.P. is given by
a - d, a, a + d, Since, Sum of the amount is Rs. 207, we have
(a - d) + a + (a + d) = 207
3a = 207 gives a = 69
It is given that product of the two least amounts is 4623.
(a - d)a = 4623 
(69 - d)69 = 4623
d = 2
Therefore, amount given by the mother to her three children are
Rs. (69 - 2), Rs. 69, Rs. (69 + 2). That is Rs. 67, Rs. 69 and Rs. 71.   

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.

L.C.M.= 3 x 2 x 2 x 2 x 5 x 3 = 360
Greatest number of 6 digit is 999999
L.C.M. of 24,15 and 36 = 360

On dividing 999999 by 360 remainder obtained is 279.
Greatest number of 6 digit, divisible by 24,15 and 36 = 999999 - 279 = 999720
Hence the required number is = 999720

By fundamental. theorem, every composite number can be expressed as a product of primes.

Factorizing 408 and 170 we get
408 = 2³ x 3¹ x 17¹
170 = 2¹ x 5¹ x 17¹
H.C.F. of 408 and 170 = 2¹ x 17¹ = 34
Also we know that H.C.F. x L.C.M.
= product of two numbers 
34 x L.C.M. = 408 x 170
L.C.M = \(\frac{408 \times 170}{34}=2040\)
H.C.F. (408, 170) = 34; L.C.M. (408, 170) = 2040

To find the H.C.F. of (340 , 412)
412 > 340 = H.C.F. (412, 340)
Using Euclid's division algorithm we have
412 = 340 x 1 + 72
The remainder 72 ≠ 0
Again applying Euclid's division algorithm
340 = 72 x 4 + 52
The remainder 52 ≠  0.
Again applying Euclid's division algorithm
72 = 52 x 1 + 20
The remainder 20 ≠  0.
Again applying Euclid's division algorithm,
52 = 20 x 2 + 12
The remainder 12 ≠  0.
Again applying Euclid's division algorithm.
20 = 12 x 1 + 8
The remainder 8 ≠ 0.
Again applying Euclid's division algorithm
12 = 8 x 1 + 4
The remainder 4 ≠  0.
Again applying Euclid's division algorithm
8 = 4 x 2+0
The remainder is 0.
Therefore H.C.F. of 340 and 412 is 4.

Since the remainders are 4, 5 respectively the required number is the HCF of the number 445 - 4 = 441, 572 - 5 = 567.
567  = 441  x 1 + 126
441 = 126 x 3 + 63 
126 = 63 x 2 + 0
Therefore HCF of 441, 567 = 63 and so the required number is 63

The number 13824 can be factorized as
2^a x 3^b = 13824 = 2⁹ x 3³
a = 9 and b = 3.

The number 800 can be factorized as
800 = 2 x 2 x 2 x 2 x 2 x 5  x 5 = 2⁵ x 5²
Hence a^b x b^a = 2⁵ x 5²
This implies that a = 2 and b = 5  or a = 5 and b = 2