Question

find the sum of 15 terms of an AP -3,-5,-7-9
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Answer 1

✯ ᴀɴsᴡᴇʀ ✯

★ given →

☞︎︎︎ AP = -3 , -5 , -7 , -9

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★ to find →

☞︎︎︎ sum of 15 terms?

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★ solution →

• a1 = first term
• d = common difference
• Sn = sum of nth terms
• n = term number

here,

$a _{1} = - 3 \\ a _{2} = - 5 \\ a _{3} = - 7$

we know that, common difference;

$d = a_{2} - a_{1} = a_{(n + 1)} - a_{n}$

here,

$d = a_{2} - a_{1} = \: a_{3} - a_{2} \\ \\ : \implies \: d = - 5 - ( - 3) = - 7 - ( - 5) \\ \\ : \implies \: d = - 5 + 3 = - 7 + 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies \: d = - 2 = - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, common difference = -2

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we know that, sum of nth terms of an AP ,S

$S = \frac{n}{2} \{ \: 2a + (n - 1)d \}$

and here,

a = -3

n = 15

d = -2

now substitute all these values in the formula,

$S_{15} = \frac{15}{2} \times \{2 \times ( - 3) + (15 - 1) \times - 2 \} \\ \\ : \implies S_{15} = \frac{15}{2} \times \{ - 6 + 14 \times ( - 2) \} \: \: \: \: \: \\ \\ : \implies S_{15} = \frac{15}{2} \times ( - 6 - 28) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies S_{15} = \frac{15}{2} \times (- 34) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies \: S_{15} = \frac{15}{ \cancel{2}} \times \cancel{ - 34} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies S_{15} = 15 \times - 17 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies S_{15} = - 255 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

hence, sum of 15 terms of AP is -255 ✔︎