Question

If the seventh term of an AP is 1/9 and its ninth term is 1/7. Find 63rd term.

Answer 1

7th term = 1/9
9th term =1/7
    we know, 
 nth term of an AP = a + (n-1)d 
              where a is the first term and d is the common difference of AP
now, 7th term = a + (7-1)d = a + 6d = 1/9
         9th term = a + (9-1)d = a +8d = 1/7
(a + 8d) - (a + 6d) = 1/7 - 1/9 = 2/63 ⇒ 2d = 2/63
          d = 1/63 
and a = 1/9 - 6d = 1/9 - 6/63 = 1/63
now,
         63th term = a + (63-1)d = a + 62d = 1/63 + 62/63 
                          = (1 + 62 )/63 = 63/63 = 1 
hence, 63th term = 1 

Answer 2

Suppose a be the first term and d be the common difference of an A.P.
So seventh term = t7 = a+(7−1)d = 19
⇒a+6d = 19......(i)
And 9th term = t9 = a+(9−1)d = 17
⇒a+8d = 17 ......(ii)
Subtracting (i) from (ii) we get;
a+8d−a−6d = 17−19⇒2d = 9−763⇒2d = 263⇒d = 163
Putting the value of d in either of the above equation , we get;
a+6×163 = 19⇒a+221 = 19⇒a = 19−221⇒a = 7−663 = 163
So 63rd term = a+(63−1)d = 163+62×163 = 1
Therefore 63rd term of an A.P. is 1.