Question

If the 7 th ​ term of an A.P. is 1/9 and 9 th term is 1/7, find 63 rd term. Thanks in advance.

Answer 1

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.

Answer 2

u can solve it
a+6d=1/9...........(1)
a+8d=1/7...........(2)
add (1)+(2)
a+6d=1/9
a+8d=1/7
-   -      -
-2d=-2/63
d=63
put value of d in (1) term and from that u find value of a and put the value of a and d in a+62d=, from this u can find the answer