Question

Which term of the sequence 25, 24 , 23 , 22,.....is the first negative term.

Answer 1

In the above Question , the following sequence is given .

25 , 24, 23, ....... .

To find -

Find the first negative term of this series .

Solution -

Given series -

=> 25 , 24, 23, 22, .......

Here , a is 25 .

The common difference , d is -1.

Now , we know that , any term of any ap series can be written as -

a_n = a + ( n - 1 ) d.

Substituting the given values -

a_n = 25 + ( n - 1 ) -1 .

Here , we have to find the value of n for which we get the minimum negative value. ..

=> 25 - n + 1 < 0

=> 26 - n < 0

=> n > 26 .

Now , the 26th term becomes -

A_26 = 25 - 25 = 0 .

So , the 27 th term of the given sequence is the first negative term .

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Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow term\:of\:sequence\:25,24,23,22,.......and\:so\:on.$

$\sf\therefore a=25$

$\sf\therefore d=t_2-t_1=24-25=-1$

$\sf\therefore d=-1$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow the\:first\:negative\:term.$

✯FORMULA IN USE,

$\large{\boxed{\bf{ \star\:\: a_n = a + ( n - 1 ) d. \:\: \star}}}$

$\large\underline\bold{SOLUTION,}$

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

$\sf\implies a_n=25+(n-1)(-1)$

$\sf\implies a_n=25-n+1$

$\sf\implies a_n=26-n$

WE KNOW,

$\sf\therefore a_n<0$

$\sf\implies 26-n>0$

$\sf\implies -n>-26$

$\sf\implies n>26$

ACCORDING TO THE QUESTION,

THE GIVEN SEQUENCES FORMS A POSITIVE TERM,

25,24,23,22,.......

THEN 26 TERM OF THE SEQUENCE WILL BE,

$\sf\therefore a_{26}= 25-25$

$\sf\implies a_{26}=0$

HENCE, THE 26TH TERM IS ZERO,

THEN THE TERMS 27,28,29,30.......AND SO ON WILL BE

NEGATIVE TERMS

$\large\underline\bold{HENCE,27^{th}\:TERM\:IS\:THE\:FIRST\:NEGATIVE\:TERM.}$ .

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