the 24th term of an AP is twice is 10th term. Show that its 72nd term is 4 times its 15 the term.

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Solution

Let first term = a

and common difference = d

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

Given that, $a_{24} = 2 \times a_{10}$

$\Rightarrow a + (24 - 1) d = 2 \times (a + (10 - 1) d)$

$\Rightarrow a + 23 d = 2 \times (a + 9 d)$

$\Rightarrow a + 23 d = 2 a + 18 d)$

$\Rightarrow a = 5 d . . . . (i)$

Now 15th term, $a_{15} = a + (15 - 1) d = a + 14 d [From (i)] = 5 d + 14 d = 19 d$

Similarly 72nd term, $a_{72} = a + (72 - 1) d = a + 71 d = 5 d + 71 d = 76 d$

Now,
$\frac{a_{72}}{a_{15}} = \frac{76 d}{19 d} = 4$

$∴ a_{72} = 4 \times a_{15}$
Hence, $72^{t h}$ term is $4$ times to its $15^{t h}$ term.

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Similar questions

24^{t h}

10^{t h}

72^{n d}

15^{t h}

The 9th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times is 15th term .

In a certian A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

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