if sum of nth term is Sn=3n²+4n finf AP and nth term
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I hope this will clear to you
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sarikajain:
By an you will found out AP also
Answered by
15
aloha!
>>-------------------☆-------------------<<
if the sum of nth term is given to be Sn=3n^2 + 4n
then by putting 1 as the value of n we get,
Sn=3n^2 + 4n
S1= 3(1)^2 + 4(1)
S1 = 3 + 4
S1 = 7
the first term is 7
now for the sum of 1st and 2nd term is:
Sn=3n^2 + 4n
S2=3(2)^2 + 4(2)
S2=3(4) + 8
S2=12 + 8
S2=20
so 2nd term = 20 - 7
= 13
now for the sum of 2nd and 3rd term:
Sn=3n^2 + 4n
S3=3(3)^2 + 4(3)
S3=3(9) + 12
S3=27 + 12
S3=39
so 3rd term= 39 - 20
= 19
so we get the A.P as follows:
7, 13, 19....
now, we have to find the nth term
for that we will first put n-1 in place of n
Sn=3n^2 + 4n
Sn-1 = 3(n-1)^2 + 4(n-1)
Sn-1= 3(n^2 + 1 - 2n) +4n -4
Sn-1= 3n^2 + 3 - 6n + 4n - 4
Sn-1= 3n^2 -2n - 1
and we know that an= Sn- Sn-1
= 3n^2 + 4n - 3n^2 +2n +1
= 6n + 1
an= 6n + 1
>>---------------☆------------------<<
hope it helps :)
>>-------------------☆-------------------<<
if the sum of nth term is given to be Sn=3n^2 + 4n
then by putting 1 as the value of n we get,
Sn=3n^2 + 4n
S1= 3(1)^2 + 4(1)
S1 = 3 + 4
S1 = 7
the first term is 7
now for the sum of 1st and 2nd term is:
Sn=3n^2 + 4n
S2=3(2)^2 + 4(2)
S2=3(4) + 8
S2=12 + 8
S2=20
so 2nd term = 20 - 7
= 13
now for the sum of 2nd and 3rd term:
Sn=3n^2 + 4n
S3=3(3)^2 + 4(3)
S3=3(9) + 12
S3=27 + 12
S3=39
so 3rd term= 39 - 20
= 19
so we get the A.P as follows:
7, 13, 19....
now, we have to find the nth term
for that we will first put n-1 in place of n
Sn=3n^2 + 4n
Sn-1 = 3(n-1)^2 + 4(n-1)
Sn-1= 3(n^2 + 1 - 2n) +4n -4
Sn-1= 3n^2 + 3 - 6n + 4n - 4
Sn-1= 3n^2 -2n - 1
and we know that an= Sn- Sn-1
= 3n^2 + 4n - 3n^2 +2n +1
= 6n + 1
an= 6n + 1
>>---------------☆------------------<<
hope it helps :)
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