Question

How many terms of the AP 43,39,35......must be taken so that their sum is 252​

Answer 1

Step-by-step explanation:

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Answer 2

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

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• First term = a = 43
• Common diff = d = -4
• Sum = 252

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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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$\sf :\implies\:{252 = \dfrac{n}{2}[2\times 43 + ( n - 1) -4]}$

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$\sf :\implies\:{252\times 2 = n[2\times 43 + ( n - 1) -4]}$

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$\sf :\implies\:{504 = n[2\times 43 + ( n - 1) -4]}$

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$\sf :\implies\:{504 = n[86 + ( n - 1) -4]}$

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$\sf :\implies\:{504 = n[86 -4n + 4 ]}$

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$\sf :\implies\:{504 = n[90 - 4n]}$

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$\sf :\implies\:{504 = 90n - 4n^2}$

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$\sf :\implies\:{504 = 2(45n - 2n^2)}$

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$\sf :\implies\:{\dfrac{504}{2} = 45n - 2n^2}$

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$\sf :\implies\:{252 = 45n - 2n^2}$

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$\sf :\implies\:{2n^2 - 45n + 252 = 0 }$

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$\sf :\implies\:{2n^2 - 24n - 21n + 252 = 0}$

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$\sf :\implies\:{2n( n - 12 ) -21 ( n - 12 ) = 0 }$

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$\sf :\implies\:{(2n - 21 ) ( n - 12 ) = 0 }$

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$\begin{tabular}{|c|c|}\cline{1-2}\sf 2n-21= 0 &\sf n - 12 = 0 \\\cline{1-2}\sf 2n = 21 &\sf n = 12 ml\\\cline{1-2}\sf n = 21 / 2 &\sf n = 12\\\cline{1-2}\end{tabular}$

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Since the no. of terms cannot be in fraction

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Therefore n = 12

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

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• 12 terms should be taken so that their sum is 252