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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.

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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...

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