There is n A.M.'s between $3$ and $17$ .The ratio of the last mean to the first mean is $3 : 1$ . Find the value of n

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Solution

$3, A_{1}, A_{2}, . . . . . A_{n}, 17$

$a_{n} = a + (n - 1) d$

$i n o u r c a s e a_{n} = 17, a = 3 a n d n u m b e r o f t e r m s = (n + 2)$

$∴ 17 = 3 + [(n + 2) - 1] d$

$17 - 3 = (n + 1) d$

$∴ d = \frac{14}{n + 1}$

$N o w, A_{1} = a + d = 3 + \frac{14}{n + 1}$

$a n d A_{n} = a + n d = a + n \frac{14}{n + 1}$

$a s g i v e n i n q u e s t i o n$
$\frac{A_{n}}{A_{1}} = \frac{3}{1}$

$O r \frac{3 + \frac{14 n}{n + 1}}{3 + \frac{14}{n + 1}} = 3$

$\frac{3 (n + 1) + 14 n}{3 (n + 1) + 14} = 3$

$\frac{17 n + 3}{3 n + 17} = 3$

$17 n + 3 = 3 (3 n + 17)$

$17 n + 3 = 9 n + 51$

$8 n = 48$

$∴ n = 6$

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