Question

how many arithmetic means are there between 10 and 34 such that second mean : last mean 3:5​

Answer 1

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Let a = 10 and b = 34

and let number of arithmetic mean between 10 and 34 be n.

So, 10, A1, A2, A3, ........., An, 34 forms an AP series.

$so \: common \: difference \: d = \frac{b - a}{n + 1} \\ = \frac{34 - 10}{n + 1} \\ = \frac{24}{n + 1}$

$\large\bold\red{according \: to \: statement}$

$\frac{A2}{An} = \frac{3}{5} \\ \frac{a + 2d}{a + nd} = \frac{3}{5} \\ 3a + 3nd = 5a + 10d \\ 2a = 3nd - 10d \\ 2 \times 10 = d(3n - 10) \\ 20 = \frac{24}{n + 1} \times (3n - 10) \\ 20n + 20 = 72n - 240 \\ 52n = 260 \\ \large\bold\red{ = > \: n = 5}$

$\huge \fcolorbox{black}{cyan}{♛Hope it helps U♛}$

Answer 2

Answer:

n = 5..................