Question

Which term of an A.P. 129, 125, 121, 117,........... is its first negative term?
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Answer 1

The required term is 32 th term.

Answer 2

Question :--- Which term of an A.P. 129, 125, 121, 117,........... is its first negative term ?

Concept used :---

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

• For T(n) be negative T(n) < 0..

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Solution :---

Here ,

a = First term = 129 .

common difference = 125 - 129 = (-4)

Since T(n) < 0.

Putting values we get,

→ 129 + (n-1)(-4) < 0

→ 129 -4n + 4 < 0

→ 133 < 4n

→ n > 133/4

→ 34 > n > 33

Since, n must be an natural number , we can say that, first negative number of AP is 34th term ..