Question

which term of the AP ; 121:117:113 ....., is the first negative term ??
( hint :- a_n<0) ​

Answer 1

QUESTION :-

which term of the AP :121,117,113 ....., is the first negative term ??

( hint :- ${a_n<0)}$

SOLUTION :-

$\leadsto$AP is 121, 117 , 113,.......

$\leadsto$the nth term of on AP is given by,

$\leadsto$ $\bold{ <strong>.</strong><strong>°</strong><strong>.</strong> \: a_n \: = a+(n-1)d}$

$\leadsto$ Here, a = 121

$\leadsto$ .°. d = 117-121 = -4

$\implies$ since, a=121 & d = -4

$\leadsto$ Let the nth term of the AP be it's first negative term.

$\leadsto$ °.° ${ a_n<0}$

$\implies$ .°. a + (n-1) d<0

________________

Now,

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

$\leadsto$ [ putting, a = 121 & d = -4)

$\leadsto$ .°. 125 + (n-1) (-4)<0

$\leadsto$ 121 - 4n + 4 < 0

$\leadsto$ 125 - 4n <0

$\leadsto$ 4n>125

$\leadsto$ n > 125/4

$\implies$ .°. n > 31.25

________________

Therefore, the first negative term is greater than 31.25 , it is 32th term.

Answer 2

Answer:

$\large\boxed{\sf{32th\;\;term}}$

Step-by-step explanation:

It's being given an AP such that,

• 121, 117, 113, .........

Here, we have,

• First term, a = 121
• Common difference, d = 117 -121 = -4

Now, we know that,

The nth term of an AP is given by,

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

Now, to find the first negative term, the nth term should be less than zero, i.e.,

$\large\bold{a_{n}<0 }$

Therefore, we have,

$= > a + (n - 1)d < 0$

Substituting the respective values, we get,

$= > 121 + (n - 1)( - 4) < 0 \\ \\ = > 121 - 4(n - 1) < 0 \\ \\ = > 121 - 4n + 4 < 0 \\ \\ = > 4n > 121 + 4 \\ \\ = > 4n > 125 \\ \\ = > n > \frac{125}{4} \\ \\ = > n > 31.25$

Thus, the first negative term will be that natural number which is just greater than 31.25.

Hence, the 32th term is first negative term .