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find the sum of invested 10% compounded annually, on which the interest for the first year plus the interest for the third year amount to $1768
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Let the sum (principal) = $100 C.I. of 1st year = \[\frac{ 100x0x1 }{ 100 } = $10\] And, amount of 1st year = $100 + $10 = $110 → The principal for 2nd year = $110 C.I. of 2nd year = \[\frac{ 100x10x1 }{ 100 } = $11\] And, amount of 2nd year = $ (110 + 11) = $121 The principal for 3rd year = $121 C.I. of 3rd year = \[\frac{ $121x10x1 }{ 100 } = $12.10\] Now, when sum of two interests = $22.10, principal (sum) = $100 And, when sum of two interests = $,768, sum = \[\frac{ $100 }{ 22.10 } x 1,768\] = $8000
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