Whic term of an A.P.:121,117,113,.......is the first negative term?
neha89:
hi
Answers
Answered by
0
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
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
do u have edit option
n>31.25 hence n should be 32
yes
i mean n>31.25
a32 = 121+(32-1)(-4) = 121 +31(-4) = 121 -124 = -4 is first negative term
It's -3 sir
Abhi , plz edit the answer
okk it's -3
how I correct it edit time over
sorry for that sir
Answered by
1
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 ;
______________________________
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 ;
______________________________
sardar ,plz change n= 31
No sir, I checked but 31th term is +1
so the answer is 32nd term which is -3
a31 = 121 +(31-1)(-4) = 121 +30(-4) = 121 -120 = 1
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