If P arithmetic means are inserted between 5 and 41 such that A3AP-1-25- where AP represents the Pth mean- then the value of P is

P A.M.'s are inserted between 5 and 41, so d = 41 5P+1 = 36P+1 Now, nbsp; A_3A_P 1=25 ⇒5+3d41 2d = 2 5 ⇒ 25+15d = 82 4d ⇒ d = 3 ⇒ nbsp;36P+1 = 3 ∴ P= 11 ...

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Standard XII

Mathematics

Sum of n Terms

If P arithmet...

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If $P$ arithmetic means are inserted between $5$ and $41$ such that $\frac{A_{3}}{A_{P - 1}} = \frac{2}{5}$ , where $A_{P}$ represents the $P^{t h}$ mean, then the value of $P$ is

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Solution

$P$ A.M.'s are inserted between $5$ and $41$ , so
$d = \frac{41 - 5}{P + 1} = \frac{36}{P + 1}$
Now,
$\frac{A_{3}}{A_{P - 1}} = \frac{2}{5} \Rightarrow \frac{5 + 3 d}{41 - 2 d} = \frac{2}{5} \Rightarrow 25 + 15 d = 82 - 4 d \Rightarrow d = 3 \Rightarrow \frac{36}{P + 1} = 3 ∴ P = 11$

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Sum of n Terms

Standard XII Mathematics

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If P arithmetic means are inserted between 5 and 41 such that A3AP−1=25, where AP represents the Pth mean, then the value of P is

P A.M.'s are inserted between 5 and 41, so d=41−5P+1=36P+1 Now, A3AP−1=25⇒5+3d41−2d=25⇒25+15d=82−4d⇒d=3⇒36P+1=3∴P=11

If $P$ arithmetic means are inserted between $5$ and $41$ such that $\frac{A_{3}}{A_{P - 1}} = \frac{2}{5}$ , where $A_{P}$ represents the $P^{t h}$ mean, then the value of $P$ is

$P$ A.M.'s are inserted between $5$ and $41$ , so
$d = \frac{41 - 5}{P + 1} = \frac{36}{P + 1}$
Now,
$\frac{A_{3}}{A_{P - 1}} = \frac{2}{5} \Rightarrow \frac{5 + 3 d}{41 - 2 d} = \frac{2}{5} \Rightarrow 25 + 15 d = 82 - 4 d \Rightarrow d = 3 \Rightarrow \frac{36}{P + 1} = 3 ∴ P = 11$