$\begin{aligned} (\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \frac{1}{10.13} + \frac{1}{13.16}) \\ is equal to = ? \end{aligned}$ $\eqalign{ & \left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right) \cr & {\text{is equal to = ?}} \cr}$

1¹/₃
2⁵/₁₆
3³/₈
4⁴¹/₇₂₈₀

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \frac{1}{10.13} + \frac{1}{13.16}) \\ Formula : \\ \frac{1}{Difference denominator value} \times [\frac{1}{First value} - \frac{1}{Last value}] \\ = \frac{1}{3} \times [1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{10} + \frac{1}{10} - \frac{1}{13} + \frac{1}{13} - \frac{1}{16}] \\ = \frac{1}{3} \times [1 - \frac{1}{16}] \\ = \frac{1}{3} \times \frac{15}{16} \\ = \frac{5}{16} \end{aligned}

$\eqalign{ & \left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right) \cr & {\text{Formula :}} \cr & \frac{1}{{{\text{Difference denominator value}}}} \times \left[ {\frac{1}{{{\text{First value}}}} - \frac{1}{{{\text{Last value}}}}} \right] \cr & = \frac{1}{3} \times \left[ {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{16}}} \right] \cr & = \frac{1}{3} \times \left[ {1 - \frac{1}{{16}}} \right] \cr & = \frac{1}{3} \times \frac{{15}}{{16}} \cr & = \frac{5}{{16}} \cr}$