The $4^{t h}$ term of an AP zero. Prove that $25^{t h}$ term of an AP is three times its $11^{t h}$ term.

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Solution

$\Rightarrow n^{t h}$ term $a_{n} = a + (n - 1) d$

Let $a_{1}$ be the first term and $d$ be the common difference of the AP.

Given, $a_{4} = 0 a_{4} = a_{1} + 3 d 0 = a_{1} + 3 d a_{1} = - 3 d (1)$

$a_{11} = a_{1} + (11 - 1) d a_{11} = a_{1} + 10 d a_{11} = - 3 d + 10 d f r o m (1) a_{11} = 7 d (2)$

$a_{25} = a_{1} + (25 - 1) d a_{25} = a_{1} + 24 d a_{25} = - 3 d + 24 d f r o m (1) a_{25} = 21 d (3)$

Dividing $(3)$ by $(2)$ $\frac{a_{25}}{a_{11}} = \frac{21 d}{7 d}$
$a_{25} = 3 a_{11}$

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