Question

if the 3rd and 9th term of AP are 4 and -8 respectively which term of the AP is zero​

Answer 1

Answer:

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Answer 2

» 3rd and 9th term of AP are 4 and -8.

Now..

$a_{n}$ = a + (n - 1)d

A.T.Q.

→ $a_{3}$ = 4

→ a + (3 - 1)d = 4

→ a + 2d = 4 _________ (eq 1)

Similarly,

→ $a_{9}$ = - 8

→ a + (9 - 1)d = - 8

→ a + 8d = - 8 ________ (eq 2)

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Subtraract (eq 1) from (eq 2)

→ a + 8d - (a + 2d) = - 8 - 4

→ a + 8d - a - 2d = - 12

→ 6d = - 12

→ d = - 2

Put value of d in (eq 1)

→ a + 2(-2) = 4

→ a - 4 = 4

→ a = 8

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We have to find an AP whose term is 0.

So,

→ $a_{n}$ = a + (n - 1)d

Put value of a and d from above calculations.

→ $a_{n}$ = 0

→ 8 + (n - 1)(-2) = 0

→ 8 + (- 2n + 2) = 0

→ 8 - 2n + 2 = 0

→ 10 - 2n = 0

→ 10 = 2n

→ 5 = n

→ n = 5

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5th term of an AP is 0.

________ [ ANSWER ]

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