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# Solved Examples on Arithmetic Progression

Illustration 1: Let a1, a2, a3, …… , a11 be real numbers satisfying a1 = 15, 27-2 a2> 0 and ak = 2ak-1 - ak-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 ak = 2ak-1 - ak-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 SP = Σai, 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 Sm/ Sndoes not depend on n. (2011)

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

Hence we have, Sm/ Sn = S5n/ Sn 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 Sm/ Sn 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.