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.

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.