Engineering
Mathematics
Arithmetic Progression and Its Properties
Question

Let a1, a2, a3.....a11 be real numbers satisfying

a1 = 15, 27 - 2a2 > 0 and ak = 2ak_1 – ak–2 for k = 3, 4..... 11.

If a12  +  a22  +.........+a11211  =  90 , then the value of  a1  +  a2  +......+  a1111  is equal to

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Solution

ak – ak – 1 = ak – 1 – ak – 1                an's are in A.P.

Let common difference d

 a12+a22+.....+a11211=152+(15+d)2+.....+(15+10d)211=990

on solving             d = – 3, – 9 (rejected) a2>272

 a1+a2+.....+a1111=112(30+10(3))11=0

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