What is the sum of \(1 \dfrac{1}{2}+ 4\dfrac{1}{6}+ 7 \dfrac{1}{12}+ 10 \dfrac{1}{20}...\) upto 20 terms?
12410/21
Integer parts 1, 4, 7, 10, … form an AP with \(a=1, d=3\). Sum of 20 terms:
\[S_I=\tfrac{20}{2}(2+19\cdot 3)=10\cdot 59=590\]
Fractional parts have denominators \(n(n+1)\), so the sum telescopes:
\[\sum_{n=1}^{20}\frac{1}{n(n+1)}=1-\frac{1}{21}=\frac{20}{21}\]
Total: \(590+\dfrac{20}{21}=\dfrac{12390+20}{21}=\dfrac{12410}{21}\).
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Important Questions from Fractions
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Column I
Column II
a.
Equivalent fraction of \(\frac{7}{12}\) is
i.
Proper fraction
b.
Equivalent fraction of \(\frac{9}{15}\) is
ii.
Improper fraction
c.
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iii.
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d.
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iv.
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