Question

how many terms of the AP 18,16,14,...,be taken so that so that their sum is zero​

Answer 1

Step-by-step explanation:

this is the correct answer

Answer 2

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

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• Sum = S = 0
• First term = a = 18
• Common diff. = d = -2

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

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• No. of terms = n = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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$\sf{\red{\boxed{\bold{S = \dfrac{n}{2} [ 2a + ( n - 1 ) d ]}}}}$

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$\sf : \implies\: {\bold{ 0 = \dfrac{n}{2} [ 2\times 18 + ( n - 1 ) - 2 ] }}$

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$\sf : \implies\: {\bold{ 0\times 2 = n [ 2\times 18 + ( n - 1 ) - 2 ] }}$

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$\sf : \implies\: {\bold{ 0 = n [ 2\times 18 + ( n - 1 ) - 2 ] }}$

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$\sf : \implies\: {\bold{ 0 = n [ 36 + ( n - 1 ) - 2 ] }}$

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$\sf : \implies\: {\bold{ 0 = n [ 36 + ( -2n + 2 ) ] }}$

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$\sf : \implies\: {\bold{ 0 = n [ 36 - 2n + 2 ] }}$

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$\sf : \implies\: {\bold{ 0 = n[ 38 - 2n ] }}$

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$\sf : \implies\: {\bold{ 0 = 38n - 2n^2}}$

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$\sf : \implies\: {\bold{ 2n^2 = 38n }}$

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$\sf : \implies\: {\bold{ n = \dfrac{38n}{2n} }}$

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$\sf : \implies\: {\bold{ n = 19 }}$

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• sum of First 19 terms of AP is zero