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Q.

If 109+2(11)1108+3(11)2(10)7+..+10(11)9=p⋅109 then p is

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a

100

b

110

c

12110

d

441100

answer is A.

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Detailed Solution

x=109+2(11)(10)8+…+10(11)9−−−1 Multiplied both sides by 11101110x=11.108+2(11)2(10)7+..+1110−−−21−2⇒x1−1110=109+11(10)8+..+119−1110                                   =109111010-11110-1-1110  sum of n terms in G.P⇒−x10=1110−1010−1110=−1010x=1011=109×100=p×109⇒p=100
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