Math, asked by Nabeela2002, 1 year ago

Whic term of an A.P.:121,117,113,.......is the first negative term?

neha89: hi

Answers

Answered by abhi178

here a=121
d=(-4)
a/c to question ,
tn <0
a+(n-1) d <0
121+(n-1)(-4) <0
121-4 (n-1)<0
121 <4 (n-1)
121/4 <(n-1)
30.25 <n-1
31.25 <n

hence n> 31.25
so the smallest n =31

mysticd: do u have edit option

SARDARshubham: n>31.25 hence n should be 32

mysticd: yes

SARDARshubham: i mean n>31.25

mysticd: a32 = 121+(32-1)(-4) = 121 +31(-4) = 121 -124 = -4 is first negative term

SARDARshubham: It's -3 sir

mysticd: Abhi , plz edit the answer

mysticd: okk it's -3

abhi178: how I correct it edit time over

abhi178: sorry for that sir

Answered by SARDARshubham

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

a = 121
d = -4

On the contrary, let 0 be the term of given A.P

an = a+(n-1)d
0 = 121 + (n-1)(-4)
n = 31.25

As the value of n is not an integer, hence 0 is not the term of given A.P ,
and the 31th term of given A.P is more than 0.

Hence the 32nd term will be the first negative term ;
______________________________

mysticd: sardar ,plz change n= 31

SARDARshubham: No sir, I checked but 31th term is +1

SARDARshubham: so the answer is 32nd term which is -3

mysticd: a31 = 121 +(31-1)(-4) = 121 +30(-4) = 121 -120 = 1

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