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Sum of the n terms of a parallel range is 3n²-n, then find the first term and the commoner
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!! Hey Mate !!
your answer is --
Given,
Sn = 3n^2-n
Now,
S1 = 3×1^2-1 = 3-1 = 2
So, first term is 2
Now,
S2 = 3×2^2 -2 = 12-2 = 10
S2 - S1 = 10-2 = 8
So, second term is 8
Now, common difference is
8 - 2 = 6
Hence, first term is 2 and common difference is 6.
【 Hope it helps you 】
your answer is --
Given,
Sn = 3n^2-n
Now,
S1 = 3×1^2-1 = 3-1 = 2
So, first term is 2
Now,
S2 = 3×2^2 -2 = 12-2 = 10
S2 - S1 = 10-2 = 8
So, second term is 8
Now, common difference is
8 - 2 = 6
Hence, first term is 2 and common difference is 6.
【 Hope it helps you 】
Anonymous:
thanks bro
welcome
bhai kuch galat hua hai kya
ha hamnay edit option dia hai edit kar lo
T1=2,T2=8 d,=6 is right answer
ya ho gya
mistake is that
mene an = 3n^2-n samjha
jbki Sn = 3n^2-n higa
ha
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