Question

If the 9th term if an ap is 99 and 99th term is 9 then find the 108th term of ap

Answer 1

$Let \:a \:and \:d \: are \: first \:term \:and \\common \: difference \: of \:an \:A.P$

$\boxed { \pink { n^{th} \:term (a_{n}) = a+(n-1)d }}$

$Given \: 9^{th} \:term = 99$

$\implies a + 8d = 99 \: ---(1)$

$Given \: 99^{th} \:term = 9$

$\implies a + 98d = 9 \: ---(2)$

/* Subtract equation (1) from equation (2) , we get */

$\implies a + 98d - (a+8d) = 9 - 99$

$\implies a + 98d - a - 8d = - 90$

$\implies 90d = - 90$

$\implies d = \frac{- 90}{90}$

$\implies d = -1$

/* Put d = -1 in equation (1), we get */

$\implies a + 8(-1) = 99$

$\implies a - 8= 99$

$\implies a = 99 + 8$

$\implies a = 107$

$Now, 108^{th} \:term \:in \:A.P \\= a + 107d \\= 107 + 107 \times (-1) \\= 107 - 107 \\ = 0$

Therefore.,

$\red { 108^{th} \:term \:in \:A.P} \green {= 0 }$

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Answer 2

Solution -:

To find -:

108 th term of AP

Required Answer -:

0

let a b the first term and DB the common difference of an ap then,

9th term T 9 = 99

$\ \ \: \tt \implies \: a \: + (9 - 1) = 99 \\ \\ \ \ \: \tt \implies \: a + 8d \: = 99$

and 9th term T99 = 9 ⟹ a + 99-1 d = 9

⟹ a + 98 d = 9

$\bold{ \:on \: subtracting \: eq \: 1 \: from \: equation \: 2 \: we \: get\: }$

⟹ 98 d 8 d = 9-99

⟹90 d = -90 ⟹ d 90/90 = -1

⟹ On putting this value in Equation we get

⟹a + 8 (-1 ) = 99 ⟹ a = 99 +8

⟹ a = 107

Now ,

108 th term ⟹ = a + (108 - 1 ) d = 107 + 107

(- 1 )

⟹ = 107 - 107

⟹ 0

thus ,, 108 th term of AP is 0 ans