Question

please answer this question ​

Answer 1

Answer:

ok

Step-by-step explanation:

djxnjxbfjxnfjddnjxxh solved!!!

Answer 2

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

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• First term = a = 6
• common difference = d = 6
• No. of times = 40

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

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• Sum of first 40 terms = ??

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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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$\sf : \implies\: {\bold{ S = \dfrac{40}{2} [ 2\times 6 + ( 40 - 1 ) 6 ] }}$

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$\sf : \implies\: {\bold{ S = \dfrac{40}{2} [ 2\times 6 + 39 \times 6] }}$

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$\sf : \implies\: {\bold{ S = \dfrac{40}{2} [ 12 + 39 \times 6] }}$

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$\sf : \implies\: {\bold{ S = \dfrac{40}{2} [ 12 + 234] }}$

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$\sf : \implies\: {\bold{ S = \dfrac{40}{2} \times 246 }}$

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$\sf : \implies\: {\bold{ S = 20 \times 123}}$

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$\sf : \implies\: {\bold{ S = 2460 }}$

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

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• Sum of first 40 multiples of 6 is 2460