Question

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❤☺If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, Find its 63rd term❤☺

Answer 1

❤❤here is your answer ✌ ✌

Suppose a be the first term and d be the common difference of an A.P.

So seventh term = t7 
⇒ a+(7−1)d = 1/9

⇒a+6d = 1/9.............(i)

And 9th term = t9 

⇒ a+(9−1)d = 1/7

⇒a+8d = 1/7 ...........(ii)

Subtracting (i) from (ii) we get:-

⇒a+8d−a−6d = 1/7−1/9
⇒2d = 9/63−7/63
⇒2d = 2/63
⇒d = 1/63

Putting 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 = 7/63−6/63 = 1/63

So 63rd term
=> a+(63−1)d 
=> 1/63+62×1/63 = 1

Therefore 63rd term of an A.P. is 1

Answer 2

Heya Mate

$\purple{\huge{\underline{\star\: Solution \: \star}}}$

seventh term = a7 = 1/9

ninth term = a9 =1/7

a7 = a + 6d

a9= a+ 8d

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.
$\pink{\huge{\underline{\star\: Answer\:1 \: \star}}}$