Number System

Let the number×be a odd no , then other no's are:
⇒ x+x+14+x+28 = 357
⇒ 3x + 42 = 357
⇒ 3x = 315
⇒ x =105
⇒ Largest number is×+28 = 105 +28 = 133
So the answer is : 133

Let the unit digit be×and ten's digit be y.
⇒ 1000y + 100x + 10y + x
= 1010y + 101x
= 101(10y + x)
So, this number is divisible by 101

Given: First term, a = -19; Last term, a_n = 36; n =12
Required sum = ($$\frac{n}{2}$$)[a + a_n]
= (12/2)[(-19 + 36)]
= 102

Tn = a + (n - 1)d
where a - first term; d - difference
Given 4th term of AP is 11,
a + (4 - 1)d = 11
a + 3d = 11
Also given that 7th term of AP is -4,
a + (7 - 1)d = -4
a + 6d = -4
After solving (1) and (2),
a = 26; d = -5
∴ 15th term = a + (15 - 1)d
= 26 + 14(-5)>= -44

Given sequence: 1, -1, 2, -2, 1, -1, 2, -2, 1,.......
In the above sequence repetition take place for each set of four terms.
Hence, remainder of 507/4 = 3.
So, that 3rd term is 2.
So that, 507th term is 2.

Given: s - 66, a - 22, l - 11
The sum of Arithmetic Progression is given by
S_n = ($$\frac{n}{2}$$)(a + l)
⇒66 = ($$\frac{n}{2}$$)(22 - 11)
⇒66 = ($$\frac{n}{2}$$)(11)
⇒n = (66×2)/11
⇒n = 12

Consider 5/8= 0.625 and 7/11 = 0.63
Now considering the options,
21/30 = 0.7
20/30 = 0.67
19/30 = 0.63
18/30 = 0.6
So, a fraction that lies between 5/8 and 7/11 is 19/30

Given:
1/6 th part painted in black and 1/3 rd part painted in red.
Remaining part painted in blue = 9m
Remaining part of the wall = 1 - (1/6 + 1/3)
= 1- [(1+2)/6]
= 1 - (3/6)
= 1 - $$\frac{1}{2}$$
= $$\frac{1}{2}$$
∴ remaining $$\frac{1}{2}$$ part painted in blue which is 9m.
So, Length of the wall = 9 + 9 = 18m.

The first number divisible by 13 = 104
The last number divisible by 13 = 390
nth term = a + (n -1 )d
Where, a = first term, d - common difference
Here, nth term = 390; a = 104; d =13
390 = 104 + (n - 1)13
(390 - 104)/13 = n - 1
286/13 = n - 1
22 = n - 1
n = 23