The 7th term of an ap is 1 upon 9 and 9th term of an ap is 1 upon 7 find its sum
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T7 = 1/9
a+6d = 1/9
a = (1/9)-6d -------------(1)
T9 = 1/7
a+8d = 1/7
a = (1/7)-8d --------------(2)
(1/9)-6d = (1/7)-8d
8d-6d = (1/7)-(1/9)
2d = 2/63
d = 1/63
a = (1/9)-6d
a = (1/9) - (6/63)
a = 1/63
T63
= a+62d
= (1/63) + 62×(1/63)
= (1/63)+(62/63)
= 63/63
= 1
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Hence the 63rd term of the given A.P is 1
Hope it helps you
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