Question

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

Answer 1

Answer:

1

Step-by-step explanation:

its answer is 1 plz mark BRAINLIEST and follow too

Answer 2

$7th\ term = T_{7} = \frac{1}{9} \\9th\ term = T_{9} = \frac{1}{7}\\T_{7} = a +6d = \frac{1}{9} \ \ \ \ ..........(i)\\T_{9} = a + 8d = \frac{1}{7}\ \ \ \ ..........(ii)\\Subtracting\ (i) from (ii) =>\\2d = \frac{1}{7} - \frac{1}{9} \\2d = \frac{9-7}{63} = \frac{2}{63}\\d = \frac{1}{63}\\\\\\ then,\ putting\ the\ value\ of\ d\ into\ (i):\\a + 6(\frac{1}{63}) = \frac{1}{9}\\a = \frac{1}{9} - 6(\frac{1}{63})\\a = \frac{7}{63} - \frac{6}{63}\\\\a = \frac{1}{63}\\then, T_{63} = a + 62d$

$= \frac{1}{63} + 62(\frac{1}{63}) \\= \frac{1}{63} + \frac{62}{63}= \frac{1+62}{63}\\= \frac{63}{63}\\= 1$

(Tn = a + (n-1)d in an AP)

Hence, the required 63rd term of the given AP is equal to 1.