# If the Seventh Term of an A.P. is 1 9 and Its Ninth Term is 1 7 , Find Its (63)Rd Term. - Mathematics

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.

#### Solution

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

We know that, nth term = an a + (n − 1)d

According to the question,

a7 =  $\frac{1}{9}$

⇒ a + (7 − 1)d = $\frac{1}{9}$

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

Also, a9 =  $\frac{1}{7}$

⇒ a + (9 − 1)d = $\frac{1}{7}$

⇒ a + 8d =  $\frac{1}{7}$    ....(2)

On Subtracting (1) from (2), we get
8− 6d =  $\frac{1}{7} - \frac{1}{9}$

⇒ 2= $\frac{9 - 7}{63}$
⇒ 2= $\frac{2}{63}$
= $\frac{1}{63}$
⇒ a = $\frac{1}{9} - \frac{6}{63}$          [From (1)]
⇒ a =   $\frac{7 - 6}{63}$
⇒ a = $\frac{1}{63}$

∴ a63 a + (63 − 1)d
=
$\frac{1}{63} + \frac{62}{63}$

= $\frac{63}{63}$   = 1

Thus, (63)rd term of the given A.P. is 1.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.4 | Q 43 | Page 26