Let us assume the number is a.
Given that 7 ¹/3 = 22/3
According to question,
a x 1/2 + a x 1/5 = a x 1/3 + 22/3
Solve the equation.

Correct Option: C

Let us assume the number is a.
Given that 7 ¹/3 = 22/3
According to question,
a x 1/2 + a x 1/5 = a x 1/3 + 22/3
⇒ a/2 + a/5 = a/3 + 22/3
⇒ a/2 + a/5 - a/3 = 22/3
⇒ (15a + 6a - 10a)/30 = 22/3
⇒ (15a + 6a - 10a) = 30 x 22/3
⇒ 11a = 10 x 22
⇒ a = 10 x 2
⇒ a = 20

Let us assume the number is n.
According to question,
If 24 is subtracted, it becomes 4/7 of the number.
⇒ n - 24 = n x 4/7
Solve the equation.

Correct Option: C

Let us assume the number is n.
According to question,
If 24 is subtracted, it becomes 4/7 of the number.
n - 24 = n x 4/7
⇒ n - 4n/7 = 24
⇒ (7n - 4n)/7 = 24
⇒ 3n/7 = 24
⇒ 3n = 24 x 7
⇒ n = 24 x 7/3
⇒ n = 8 x 7
⇒ n = 56
Sum of the digits of the number = 5 + 6 = 11

Let us assume the first term is a and common difference is d.
Formula for n^th term of Arithmetic Progression = a + ( n - 1 ) x d
According to question,
Sum of the first term and the fifth term = 10
t₁ + t₅ = 10

Formula of sum of n terms for Arithmetic Progression,
S_n = n/2[ 2a + ( n - 1 ) x d ]

Again According to question,
The sum of all terms of the Arithmetic Progression, except for the 1st term = 99
S₁₀ - first term = 99

Solve the equation.

Correct Option: B

Let us assume the first term is a and common difference is d.
Formula for n^th term of Arithmetic Progression = a + ( n - 1 ) x d
So first term t₁ = a + ( n - 1 ) x d
t₁ = a + (1 - 1) x d = a + 0 x d = a
Fifth term t₅ = a + ( n - 1 ) x d = a + (5 - 1) x d = a + 4d
According to question,
Sum of the first term and the fifth term = 10
t₁ + t₅ = 10
⇒ a + a + 4d = 10
⇒ 2a + 4d = 10
⇒ a + 2d = 5.............................(1)

Formula of sum of n terms for Arithmetic Progression,
S_n = n/2[ 2a + ( n - 1 ) x d ]
Put the value of n = 10, Since total number of terms is 10 .
⇒ S₁₀ = 10/2[ 2a + ( 10 - 1 ) x d ]
⇒ S₁₀ = 5[ 2a + 9 x d ] = 10a + 45d
Again According to question,
The sum of all terms of the Arithmetic Progression, except for the 1st term = 99
S₁₀ - first term = 99
⇒ 10a + 45d - a = 99
⇒ 9a +45d = 99
⇒ a + 5d = 11.................................(2)
Subtracts Equation (1) from Equation (2), we will get,
⇒ a + 5d - a - 2d = 11 - 5
⇒ 3d = 6
⇒ d = 2
Put the value of d in equation (1), we will get
⇒ a + 2 x 2 = 5
⇒ a + 4 = 5
⇒ a = 5 - 4
⇒ a = 1
Since t_n = a + ( n - 1) x d
find the third term by putting the value of a , n and d. we will get,
t₃ = 1 + ( 3 - 1) x 2 = 1 + 2 x 2 = 1 + 4 = 5
So 3rd term of Arithmetic Progression is 5.

let us assume a be the first term and d the common difference.
According to Question,
m^th term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for m^th term,
⇒ a + ( m - 1 ) x d = t_m
Use the Arithmetic Progression formula for n^th term,
⇒ a + ( n - 1 ) x d = t_n
Now according to question,
Sum of mn terms = mn/2(2a + ( mn - 1) x d)
S_mn = mn/2(2a + ( mn - 1) x d)

Solve the equation

Correct Option: C

let us assume a be the first term and d the common difference.
According to Question,
m^th term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for m^th term,
⇒ a + ( m - 1 ) x d = t_m
⇒ a + ( m - 1 ) x d = 1/n
⇒ a + md - d = 1/n .......................(1)
Use the Arithmetic Progression formula for n^th term,
⇒ a + ( n - 1 ) x d = t_n
⇒ a + ( n - 1 ) x d = 1/m
⇒ a + nd - d = 1/m.........................(2)
Subtracts Equation (1) from Equation (2) , we will get
⇒ a + nd - d - (a + md - d) = 1/m - 1/n
⇒ a + nd - d - (a + md - d) = 1/m - 1/n
⇒ a + nd - d - a - md + d = 1/m - 1/n
⇒ nd - md = 1/m - 1/n
⇒ d(n - m) = 1/m - 1/n
⇒ d(n - m) = (n - m)/mn
⇒ d = 1/mn.......................................(3)
Put the value of d in Equation (1), we will get,
⇒ a + md - d = 1/n
⇒ a + ( m x 1/mn ) - 1/mn = 1/n
⇒ a + 1/n - 1/mn = 1/n
⇒ a = 1/n - 1/n + 1/mn
⇒ a = 1/mn.....................................(4)
Now according to question,
Sum of mn terms = mn/2(2a + ( mn - 1) x d)
S_mn = mn/2(2a + ( mn - 1) x d)
Put the value of a and d in above equation, we will get,
S_mn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ]
S_mn = mn/2 [2/mn + 1 - 1/mn ]
S_mn = mn/2 [ 1 + 1/mn ]
S_mn = mn/2 [ ( mn + 1 )/mn ]
S_mn = mn/2 (1 + mn)/mn
S_mn = 1/2 (1 + mn)
S_mn = (1 + mn)/2 = (mn + 1)/2

According to question,
First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10.
Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94.
The common difference between two consecutive numbers will be 7.
This series will be like →10, 17, 21 ,............................ 94.
Here First Number a = 10, Last number t_n = 94, Common Difference d = 7 ;
Using the Formula for Last Number t_n = a + (n - 1) x d;
⇒ a + (n - 1) x d = t_n

Using the formula for the sum of Arithmetic Progression.
S_n = n/2 [ First Number + Last Number ] ;
Put the value of n , First Number and Last Number.
Solve the equation.

Correct Option: B

According to question,
First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10.
Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94.
The common difference between two consecutive numbers will be 7.
This series will be like →10, 17, 21 ,............................ 94.
Here First Number a = 10, Last number t_n = 94, Common Difference d = 7 ;
Using the Formula for Last Number t_n = a + (n - 1) x d;
⇒ a + (n - 1) x d = t_n
Put the value of a , d and t_n in above equation.
⇒ 10 + (n - 1) x 7 = 94
⇒ 10 + 7n - 7 = 94
⇒ 7n + 3 = 94
⇒ 7n = 94 - 3 = 91
⇒ n = 91/7
⇒ n = 13
Using the formula for the sum of Arithmetic Progression.
S_n = n/2 [ First Number + Last Number ] ;
Put the value of n , First Number and Last Number, we will get,
S_n = 13/2 [ 10 + 94 ]
S_n = 104 x 13/2
S_n = 52 x 13
S_n = 676