Answers

2016-02-24T18:48:39+05:30
Suppose a be the first term and d be the common difference of an A.P.So seventh term = t 7 = a + 7 - 1 d = 1 9⇒ a + 6 d = 1 9 . . . . . . ( i )And 9th term = t 9 = a + 9 - 1 d = 1 7⇒ a + 8 d = 1 7 . . . . . . ( ii )Subtracting (i) from (ii) we get;a + 8 d - a - 6 d = 1 7 - 1 9 ⇒ 2 d = 9 - 7 63 ⇒ 2 d = 2 63 ⇒ d = 1 63Putting the value of d in either of the above equation , we get;a + 6 × 1 63 = 1 9 ⇒ a + 2 21 = 1 9 ⇒ a = 1 9 - 2 21 ⇒ a = 7 - 6 63 = 1 63So 63rd term = a + 63 - 1 d = 1 63 + 62 × 1 63 = 1Therefore 63rd term of an A.P. is 1.
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2016-02-24T19:37:52+05:30
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.



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