Question

Which term of ap call on 121, 117, 113, 118 is its first negative term?

Answer 1

Answer: 32!

Step-by-step explanation:

Given AP is:

121, 117, 113, ................

First term a = 121

Common difference d = 117-121 = -4

Let nth term is the first negative term in AP

Now nth tern in AP = a + (n-1)*d

= 121 + (n - 1)*(-4)

= 121 - 4n + 4

= 125 - 4n

Now, we have to find the suitable value of n so that the value of (125 - 4n) is negative.

=> 125 - 4n < 0

=> 125 < 4n

=> 4n > 125

=> n > 125/4

=> n > 31.25

Since the term can not be in decimal form (term is always taken a positive integer)

So, n = 32

Now, by taking value of n = 32, we get

nth term = 125 - 4*32

= 125 - 128

= -3

This is the first negative term in the series.

So, n = 32

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

HOPE IT HELPS YOU!

Answer 2

$\huge \bf{Question}$

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

$\huge \bf{Solution :-}$

Given, AP is 121, 117, 113, ... .

Here, first term, a = 121

and common difference, d = 117 – 121 = – 4

Let, the nth term of this AP be the first negative

term. Then,

$\boxed{ \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: a_n \: < 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

$\boxed{\implies \: \: \bf a + (n - 1) \: d < 0 \implies \: 121 + (n - 1)( - 4) < 0 \: }$

$\boxed{ \implies \: \: \bf 125 - 4n < 0 \implies \: 125 < 4 n \implies \: 4n > 125}$

$\boxed{ \implies \: \: \bf n > \frac{125}{4} \implies \: n > 31 \frac{1}{4} \: \: \: \: \: \: \: \: \: \: }$

$\sf \: So, \: least \: integral \: value \: of \: n \: is \: 32.$

$\sf \: Hence, \: \pink{\underline{ \underline{32nd}}} \: term \: of \: the \: given \: AP \: is \: the \: first \: negative \: term.$