Q. 14.3( 228 Votes )

# Which term of the AP: 121, 117, 113, . . ., isits first negative term?

[Hint: Find *n* for ]

Answer :

Here,

first term, a = 121

common difference, d = a

_{2}- a

_{1}= 117 - 121 = -4

Let the first negative term be 'a

_{n}'.

Also, we know that, nth term of an AP is given by

a

_{n}= a + (n - 1)d

We have to find least value of n, such that

a

_{n}< 0

⇒ a + (n - 1)d < 0

⇒ 121 + (n - 1)(-4) < 0

⇒ 121 - 4(n - 1) < 0

⇒ 4(n - 1) > 121

⇒ 4n - 4 > 121

⇒ 4n > 125

⇒ n > 31.25

Therefore, n is 32 [least positive integer greater than 31.25 is 32]

Hence, the 32nd term of AP is first negative term.

Also,

a

_{32}= a + 31d

= 121 + 31(-4)

= 121 - 124 = -3

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