Question

write the sum of first 7 terms of the arithmetic sequence 1/7, 2/7, 3/7
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Answer 1

Step-by-step explanation:

$\frac{1}{7}, \: \frac{2}{7}, \: \frac{3}{7}, \: ................., \: \frac{7}{7} \\ a = \frac{1}{7} \\n = 7 \\ d = \frac{2}{7} - \frac{1}{7} \\ = \frac{1}{7} \\ \\ Sn = \frac{n}{2} (2a + (n - 1)d) \\ = \frac{7}{2} (2 \times \frac{1}{7} + (7 - 1) \frac{1}{7} ) \\ = \frac{7}{2} ( \frac{2}{7} + (6 \times \frac{1}{7} )) \\ = \frac{7}{2} ( \frac{2}{7} + \frac{6}{7} ) \\ = \frac{7}{2} \times \frac{8}{7} \\ = \frac{8}{2} \\ = 4$

Answer 2

$\large\underline{ \underline{ \sf{ \ddag \qquad Your\:question: \qquad \ddag}}}$

$\:$

$\bigstar$What is the sum of the first 7 terms of the arithmetic sequence 1/7, 2/7, 3/7.

$\:$

$\large\underline{ \underline{ \sf{ \ddag \qquad Given: \qquad \ddag}}}$

$\\ \hookrightarrow \tt \: a = \frac{1}{7}$

$\\ \hookrightarrow \tt \: n = 7$

$\large\underline{ \underline{ \sf{ \ddag \qquad Solution:\qquad \ddag}}}$

$\bigstar$First we have to know the common difference.

$\\ \hookrightarrow \tt \: d = \left( \frac{2}{7} - \frac{1}{7} \right)$

$\\ \hookrightarrow \tt \: d = \left( \frac{1}{7} \right)$

$\bigstar$To find the sum of first 7 terms using formula,

$\\ \hookrightarrow{ \underline{ \boxed{ \tt{ \pink{S_n = \frac{n}{2} \times \left(2a + (n - 1) \times d \right) }}}}} \star$

$\\ \hookrightarrow\tt{S_n = \frac{7}{2} \times \left(2 \times \left( \frac{1}{7} \right) + \left[7 - 1 \right]\times \frac{1}{7} \right) }$

$\\ \hookrightarrow\tt{S_n = \frac{7}{2} \times \left( \frac{2}{7} + \left[7 - 1 \right]\times \frac{1}{7} \right) }$

$\\ \hookrightarrow\tt{S_n = \frac{7}{2} \times \left( \frac{2}{7} + \frac{6 \times 1}{7} \right)}$

$\\ \hookrightarrow\tt{S_n = \left(\frac{7}{2} \times \frac{8}{7} \right) }$

$\\ \hookrightarrow{ \underline{ \boxed{\tt{ \pink{S_n =4 }}}}} \star$

$\bigstar$Hence,The sum of 7 th term is 4.