Let a = 1 1 1....1 (55 digits), b=1 +10+10 $^{2}$ +....+10 $^{4}$ , c =1+10 $^{5}$ +10 $^{10}$ +10 $^{15}$ +....+10 $^{50}$ , then

a =b+c

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a =bc

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b= ac

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c =ab

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Solution

The correct option is B
a =bc

$a = 1 + 10 + 10^{2} + . . . . + 10^{54}$

$= \frac{10^{55} - 1}{10 - 1} = \frac{10^{55} - 1}{10^{5} - 1} \times \frac{10^{5} - 1}{10 - 1} = b c$

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Similar questions

b = 1 + 10 + 10^{2} + 10^{3} + 10^{4}

c = 1 + 10^{5} + 10^{10} + 10^{15} + . . . + 10^{50}, t h e n

a = 111.........1      55, b = 1 + 10 + 10^{2} + 10^{3} + 10^{4}

c = 1 + 10^{5} + 10^{10} + . . . . + 10^{50}

a i s n o t e q u a l b c

10^{- 1} \times 10^{2} \times 10^{- 3} \times 10^{4} \times 10^{5} \times 10^{6}

Find the number from each of the following expanded forms:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰

(b) 4 × 10⁵+ 5 × 10³+ 3 × 10²+ 2 × 10⁰

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹

$(\frac{1}{10}) + {(\frac{1}{10})}^{2} + {(\frac{1}{10})}^{3} + {(\frac{1}{10})}^{4} =$ ?

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