Question

which term of AP:121,117,113,......is it's first term negative (Hint:Find n for an
<0)​

Answer 1

Answer:

General term or nth term of A.P

The general term or nth term of A.P is given by an = a + (n – 1)d,

where a is the first term, d is the common difference and n is the number of term.

----------------------------------------------------------------------------------------------------

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. 

----------------------------------------------------------------------------------------------------

Hope this will help you...

Answer 2

$\rule{200}3$

$\Huge{\red{\underline{\textsf{Answer}}}}$

$\sf AP\: is \: 121,\: 117,\: 113,......$

$\sf The\:n^{th}\: term \:of \: on \: AP\: is given\: by$

$\sf a_n = a + (n - 1)d$

$\rule{200}3$

$\large{\pink{\underline{\tt{Given :-}}}}$

$\sf a = 121$

$\sf d = 117 - 121 = -4$

$\rule{200}3$

$\large{\orange{\underline{\tt{Solution :-}}}}$

Let the nth term of the AP be the first negative term.

$\sf a_n < 0$

$\sf a + (n - 1)d < 0$

By putting a = 121 and d = -4

$\purple\longrightarrow$ $\purple{\sf 121 + (n - 1) (-4) < 0 }$

$\purple\longrightarrow$ $\purple{\sf 121 - 4n + 4 < 0}$

$\purple\longrightarrow$ $\purple{\sf 125 - 4n < 0}$

$\purple\longrightarrow$ $\purple{\sf 125 > 4n}$

$\purple\longrightarrow$ $\purple{\sf 4n < 125}$

$\purple\longrightarrow$ $\purple{\large\sf n > \frac{125}{4}}$

$\pink\longrightarrow$ $\underline{\boxed{\pink{\sf n > 31.25}}}$ $\orange\star$

Since the negative term is greater than 31.25, it is 32nd term.

$\rule{200}3$