Question

Which term of the AP : 121, 117, 113, . . ., is its first negative term?​

Answer 1

Answer:

PLEASE MARK IT AS BRAINLIEST

Step-by-step explanation:

Given:first term(a)= 121 

common difference (d)= 117- 121 = -4

∵ n th term of an AP

an = a + (n – 1)d

⇒121+(n-1) ×(-4)

⇒121-4n+4

⇒12+4-4n

⇒125 -4n

an= 125 -4n

For first negative term , an <0

⇒ 125-4n<0

⇒125<4n

⇒4n>125

⇒n>125/4

⇒n> 31 1/4

least integral value of n= 32

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

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Hope this will help you...

Answer 2

Answer:

n = 32

Step-by-step explanation:

Let the $\sf n^{th}$ term be the first negative term of the AP.

The AP is, 121, 117, 113, . . . Here ;

• a = 121
• d = 117 - 121 = - 4

Clearly, the first negative term will come after zero. Consider, the value of $\sf a_n$ to be less than zero.

$\sf \implies a_n < 0 \\\\\implies \sf a + (n-1)d <0 \\\\\implies \sf 121 + (n-1)(-4) < 0 \\\\\implies \sf 121 - 4(n - 1) < 0\\\\\implies \sf 121 - 4n +4 < 0 \\\\\implies \sf 125 - 4n < 0 \\\\\implies \sf 4n > 125 \\\\\implies \sf n > \frac{125}{4} \\\\\implies \sf n > 31.2$

The possible value of n is 32.

Hence, the $\bf 32^{th}$ term of the AP is the first negative term.