Question

If the seventh term of an AP is
$\frac{1}{9}$
And its ninth term is

$\frac{1}{7}$
Find its 63rd term

Answer 1

Answer :

Let, the first term of the AP be a and the common difference be d

Then, 7th term = a + (7 - 1) d = a + 6d

and 9th term = a + (9 - 1) d = a + 8d

By the given conditions,

a + 6d = $\frac{1}{9}$ ...(i)

a + 9d = $\frac{1}{7}$ ...(ii)

Now, (ii) - (i) ⇒

a + 9d - a - 6d = $\frac{1}{7} - \frac{1}{9}$

⇒ 3d = $\frac{9 - 7}{63}$

⇒ 3d = $\frac{2}{63}$

⇒ d = $\frac{2}{189}$

Now, putting d = $\frac{2}{189}$ in (i), we get

a + $6 ( \frac{2}{189}) = \frac{1}{9}$

⇒ a + $\frac{12}{189} = \frac{1}{9}$

⇒ a = $\frac{1}{9} - \frac{12}{189}$

⇒ a = $\frac{21 - 12}{189}$

⇒ a = $\frac{9}{189}$

⇒ a = $\frac{1}{21}$

Hence, the 63rd term of the AP is

= a + (63 - 1) d

= a + 62d

$= \frac{1}{21} + 62 (\frac{2}{189})$

$= \frac{9 + 124}{189}$

$= \frac{133}{189}$

$= \bold{\frac{19}{27}}$

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