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Question

Let a1,a2,...,a10 be an AP with common difference 3 and b1,b2,...,b10 be a GP with common ratio 2. Let ck=ak+bk, k=1,2,...,10. If c2=12 and c3=13, then 10k=1ck is equal to

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Solution

c2=a2+b2=(a13)+2b1=12a1=11
c3=a3+b3=(a16)+4b1=13b1=2
ck=ak+bk=(a13(k1))+(b12k1)
=(113k+3)+(2k)=143k+2k
10k=1 ck=10k=1(2k3k+14)
=10k=12k310k=1k+10k=114
=2(2101)310112+140=2021

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