The Sum of all terms of the arithmetic progression having 10th terms except for the 1st term, is 99, and except for the 6th term , 89. Find the 3rd terms of the progression if the sum of the 1st and the 5th term is equal to 10.

Required measure = HCF of 403, 434 and 465 = 31 cm

Length of to the rope = Radius of circle
According to the question,
?r² = 154
? r² = 154 x (7/22) = 7 x 7 = 49
? r = ?49 = 7 m

Let CP = x
According to the question,
625 - x = x - 435
? 2x = 1060
? x = 1060/2 = ? 530

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.

The difference of requisite number must be 12 and each one must be divisible by 12, so the numbers are 84, 96.

The largest number is the H.C.F. of 210, 315, 147, and 168, which is 21.

Suppose there are 8y questions were asked apart from the 41 question.Then
37 + 5y/41 + 8y = 80% = 4/5
? 185 + 25y = 164 + 32y
? 7y = 21
? y = 3
? Total no. of questions = 41 + 8 x 3 = 65.

Given, a = 20%
Required percentage = [a/(100 - a)] x 100%
= [20/(100 - 20)] x 100%
= (20/80) x 100%
= 25%

? 5% of A=15% of B and 10% of B = 20% of C
? A/20 = 3B/20 and B/10 = C/5
? B = 2C
? A/20 = (3/20) x 2C = 3C/10 =(3/10) x 2000 = 600
? A = (600 x 20) = 12000,
B = (2 x 2000) = 4000
? A + B + C = (12000 + 4000 + 2000) = 18000

( A + B) 's 1 days work = 1/8
B's 1 days work = 1/12
? A's 1 day's work = (1/8) - (1/12) = (3 - 2)/24 = 1/24
? A can complete the work in 24 days
Now, B's 4 days work = 4/12 = 1/3
Remaining work = 1 - 1/3 = 2/3
As, time taken by A to complete the work is 24 days.
? Time taken by A to do 2/3 of the work = 2/3 x 24 = 16 days