Solved Examples on Arithmetic Progression 

Illustration 1: Let a₁, a₂, a₃, …… , a₁₁ be real numbers satisfying a₁ = 15, 27-2 a₂> 0 and a_k = 2a_k-1 - a_k-2 for k = 3, 4, …. , 11. If (a₁² + a₂² + a₃² + …… + a₁₁²)/11 = 90, then find the value of (a₁ + a₂ + a₃ + …… + a₁₁)/11.(2010)

Solution: Given that a_k = 2a_k-1 - a_k-2

This relation clearly implies that a₁, a₂, a₃, …… , a₁₁ are in A.P.

Hence, (a₁² + a₂² + a₃² + …… + a₁₁²)/11 = [11a² + 35.11d² +10ad]/11 = 90.

Hence, we obtain, 225 + 35.d² +150d = 90

So, 35.d² +150d + 135 = 0

This gives d = -3, -9/7

But, it is given that a₂< 27/2 and hence, d = -3, d ≠ -9/7.

So, (a₁ + a₂ + a₃ + …… + a₁₁)/11 = 11/2 [30-10.3] = 0.

Illustration 2: Let a₁, a₂, a₃, …… , a₁₀₀ be an arithmetic progression with a₁ = 3 and S_P = Σa_i, where the summation runs over p and 1 ≤ p ≤ 100. For any integer n with 1 ≤ n ≤ 20, let m = 5n. What is the value of a₂ if S_m/ S_ndoes not depend on n. (2011)

Solution: It is given in the question thata₁ = 3 and m = 5n.

Hence we have, S_m/ S_n = S_5n/ S_n is independent of n

So, if we have 6-d = 0 then it gives d = 6.

So, a₂ = a₁ + d = 9

Or

if d = 0,  then S_m/ S_n is independent of n.

This gives the value of a₂ = 3.

Illustration 3: The fourth power of the common difference of an A.P. with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer. (2000)

Solution:Let four consecutive terms of the A.P. a-3d, a-d, a+d, a+3d

Then the product is given by

P = (a-3d)(a-d)(a+d)(a+3d) + (2d)⁴

   = (a²-9d²)(a²-d²) + 16d⁴

   = (a²-5d²)²

Now, (a²-5d²) = a²-9d² + 4d²

(a-3d)(a+3d) + (2d)²

= I.I + I²

= I

Therefore, P = I²= Integer.