-
Courses-
Crash Courses 2021
-
Olympiads/NTSE/Others
-
Micro Courses
-
Vedantu Super Kids NEW
-
Reading & Spoken English -
FREEMaster Classes
-
PREMIUM1-on-1 Classes
-
Study Material
-
NCERT Solutions
-
Previous Year Papers
-
Mock Tests
-
Sample Papers
-
Reference Book Solutions
-
ICSE Solutions
-
School Syllabus
-
Revision Notes
-
Important Questions
-
Math Formula Sheets
-
-
More
Book your FREE Online counselling session
Share your contact information
Your FREE Online counselling session has been booked!
Vedantu academic counsellor will be calling you shortly for your Online Counselling session.
DONE
Please try again later
OK
Questions & Answers
Question
Answers
Related Questions
Which term of the given A.P. 121, 117, 113,….. is the first negative term?
Answer
![Verified]()
Verified
Verified
Hint: Start with finding out the general term of the A.P. Then put it less than zero to find the first negative term.
According to the question, the given sequence is 121, 117, 113,…..
As we can see, the first term of A.P. is 121 and its common difference is -4. So, we have:
$ \Rightarrow a = 121{\text{ and }}d = - 4$
We know that the general term of A.P. is:
$ \Rightarrow {T_n} = a + \left( {n - 1} \right)d$
Putting the values of $a$ and $d$ from above, we get:
$
\Rightarrow {T_n} = 121 + \left( {n - 1} \right)\left( { - 4} \right), \\
\Rightarrow {T_n} = 121 - 4n + 4, \\
\Rightarrow {T_n} = 125 - 4n \\
$
For the term in the sequence to be negative, we have:
$
\Rightarrow {T_n} < 0, \\
\Rightarrow 125 - 4n < 0, \\
\Rightarrow 4n > 125, \\
\Rightarrow n > \dfrac{{125}}{4}, \\
\Rightarrow n > 31.25 \\
$
Therefore, the first negative term of A.P. is the 32nd term.
Note:
$n = 31$ will be the last positive term of the given A.P. and $n \ne 31.25$. $n$ must be an integer. So, $n = 32$ will be the first negative term.
According to the question, the given sequence is 121, 117, 113,…..
As we can see, the first term of A.P. is 121 and its common difference is -4. So, we have:
$ \Rightarrow a = 121{\text{ and }}d = - 4$
We know that the general term of A.P. is:
$ \Rightarrow {T_n} = a + \left( {n - 1} \right)d$
Putting the values of $a$ and $d$ from above, we get:
$
\Rightarrow {T_n} = 121 + \left( {n - 1} \right)\left( { - 4} \right), \\
\Rightarrow {T_n} = 121 - 4n + 4, \\
\Rightarrow {T_n} = 125 - 4n \\
$
For the term in the sequence to be negative, we have:
$
\Rightarrow {T_n} < 0, \\
\Rightarrow 125 - 4n < 0, \\
\Rightarrow 4n > 125, \\
\Rightarrow n > \dfrac{{125}}{4}, \\
\Rightarrow n > 31.25 \\
$
Therefore, the first negative term of A.P. is the 32nd term.
Note:
$n = 31$ will be the last positive term of the given A.P. and $n \ne 31.25$. $n$ must be an integer. So, $n = 32$ will be the first negative term.
×
Sorry!, This page is not available for now to bookmark.
Like this
Liked
NCERT Book Solutions
Reference Book Solutions
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
(Mon-Sun: 9am - 11pm IST)
Questions?
Chat with us
