Math, asked by uday3104, 1 year ago

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Answered by Bhasksr
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GIVEN:

 {n}^{th} term \:of \: an \: arithmetic \: progression = n + 5 \\ putting \: n = 1 \\ 1st \: term \: a \: = 6 \\ putting \: n = 2 \: \\ sum \: of \: 1st \: and \: 2nd \: terms = 7 \\ 2nd \: term \: = 7 - 6 = 1 \\ common \: diff = -5\\ therefore \: sum \: of \: first \: 10terms \: are \\ s_{10} = \frac{10}{2} (2 \times 6 + 9 \times -5) \\ = > s_{n} = -165 \\

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