Question

how many numbers of the series 148 + 146 + 144 + 142 and so on must be taken to havesum zero

Answer 1

Answer:

70 (i guess

Step-by-step explanation:.........

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

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

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• An AP : 148 + 146 + 144 + 142.......

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$\red{\underline \bold{To \: Find:}}$

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• Number of terms required so that the some becomes 0.

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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We know that,

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$\purple\longrightarrow$ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$

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• a = 148

• d = -2

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$\sf \longmapsto S_n = \dfrac { n } { 2 } (2(148)+ (n - 1)(-2))$

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$\sf \longmapsto S_n = \dfrac { n } { 2 } (296 + -2n +2)$

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$\sf \longmapsto S_n = \dfrac { n } { 2 } (296 + -2n +2)$

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$\sf \longmapsto 0 = \dfrac { n } { 2 } (298 -2n )$

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$\sf \longmapsto 0 = (298 + -2n )$

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$\sf \longmapsto 0 = 149 - n$

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$\sf \longmapsto n = 149$

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Hence 149 terms will add up so that the sum becomes 0

$\rule{200}5$