Q.

For any sequence of real number  A={a1,a2,a3,.....}, a sequence  ΔA  is defined such that  ΔA={a2a1,a3a2,a4a3,.....}. Suppose that all of the terms of the sequence Δ(ΔA)  are 1 and that  a19=a92=0. The units digit in  a4 is _____.

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Detailed Solution

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Suppose that the first term of the sequence  ΔAis  d  then
 ΔA={d,d+1,d+2,.....(d+(n1))}
Hence  A=(a1,a1+d,a1+d+(d+1),a1+d+(d+1)+(d+2)......)
a0=a1+(n1)d+12(n1)(n2) 
So  an is a quadratic polynomial in n.
So,  an=12(n19)(n92)        So,  a3=12(319)(392)=712              a4=12(419)(492)=12(15)(88)=(44)(15)=660

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