Answer in Brief

If the seventh term of an A.P. is \[\frac{1}{9}\] and its ninth term is \[\frac{1}{7}\] , find its (63)^{rd} term.

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#### Solution

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{7} = \[\frac{1}{9}\]

⇒ *a* + (7 − 1)*d* = \[\frac{1}{9}\]

⇒ *a* + 6*d* = \[\frac{1}{9}\] * ....* (1)

Also, *a*_{9} = \[\frac{1}{7}\]

⇒ *a* + (9 − 1)*d* = \[\frac{1}{7}\]

⇒ *a* + 8*d* = \[\frac{1}{7}\] ....(2)

On Subtracting (1) from (2), we get

8*d *− 6*d* = \[\frac{1}{7} - \frac{1}{9}\]

⇒ 2

*d*= \[\frac{9 - 7}{63}\]⇒ 2

*d*= \[\frac{2}{63}\]⇒

⇒

⇒

Thus, (63)

*d*= \[\frac{1}{63}\]⇒

*a*= \[\frac{1}{9} - \frac{6}{63}\] [From (1)]⇒

*a*= \[\frac{7 - 6}{63}\]⇒

*a*= \[\frac{1}{63}\]∴

= \[\frac{63}{63}\] = 1*a*_{63 }=*a*+ (63 − 1)*d*

=\[\frac{1}{63} + \frac{62}{63}\]=

Thus, (63)

^{rd}term of the given A.P. is 1.

Concept: Sum of First n Terms of an AP

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