Sequences and Series 4 Question 5

5. If (10)9+2(11)1(10)8+3(11)2(10)7++10(11)9=k(10)9, then k is equal to

(2014 Main)

(a) 12110

(b) 441100

(c) 100

(d) 110

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Solution:

  1. Given,

k109=109+2(11)1(10)8+3(11)2(10)7++10(11)9k=1+21110+311102++1011101110k=11110+211102++911109+101110

On subtracting Eq. (ii) from Eq. (i), we get

k101110=111101011110110111010

In GP, sum of n terms =a(rn1)r1, when r>1

k=101011101010101110k=100



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