Question

If seventh term of AP is 1/9 and it 's ninth term is 1/7 find its 63rd term

Answer 1

nth term = a+ ( n-1 ) d
( where a is the first term, n is the no. of terms
and d is the difference between two consecutive terms . )

seventh term = a7 = 1/9
ninth term = a9 =1/7
a7 = a + 6d - ( 1 )
a9= a+ 8d - ( 2 )
by subtracting ( 1 ) from ( 2 )
(+ )a+ 8d = 1/7 ( +)
( - )a+ 6d = 1/9 ( - )
= 2d = 2 /63
d = 1/63
by putting value of d in eq. ( 2 )
a + 8 ( 1/63 ) = 1/7
a + 8/63 = 1/7
a = 1/7 - 8/63
a = 1/63

a 63 = a + 62d
=1/63 + 62 ( 1 /63 )
= 1/63 + 62/63
= 1 ans.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the first term be a and the common difference be d of the given A.P.

Given, a(7) = 1/9

⇒ a + 6d = 1/9 ..... (i)

⇒ a(9) = 1/7

⇒ a + 8d = 1/7 .... (ii)

On subtracting eq (i) from (ii), we get

⇒ a + 8d - a - 6d = 1/7 - 1/9

⇒ 2d = 2/63

⇒ d = 1/63

Substituting the value of d in Eq (ii), we get

⇒ a + 8 × 1/63 = 1/7

⇒ a = 1/7 - 8/63

⇒ a = 9 - 8/63

⇒ a = 1/63

Now, a(63) = 1/63 + 62 × 1/63

⇒ a(63 = 1 + 62/63

⇒ a(63) = 63/63

⇒ a(63) = 1

Hence, the 63rd term of an A.P. is 1.