Asked by blehhh.binga
The sum of the first n terms of a progression is Sn where Sn=2nsquared - n. Prove that the progression is an AP, stating the first term and the common difference.
Answers
Answered by
oobleck
Sn = 2n^2 - n
so just plug that into the AP formula and you have
n/2 (2a + (n-1)d) = 2n^2 - n
an + dn^2/2 - dn/2 = 2n^2 - n
d/2 n^2 + (a - d/2)n = 2n^2 - n
equating coefficients, we get
d/2 = 2
a - d/2 = -1
leaving us with
a = 1
d = 4
check:
n/2 (2 + 4(n-1)) = n/2 (2 + 4n - 4) = n/2 (4n-2) = 2n^2-n ✅
so just plug that into the AP formula and you have
n/2 (2a + (n-1)d) = 2n^2 - n
an + dn^2/2 - dn/2 = 2n^2 - n
d/2 n^2 + (a - d/2)n = 2n^2 - n
equating coefficients, we get
d/2 = 2
a - d/2 = -1
leaving us with
a = 1
d = 4
check:
n/2 (2 + 4(n-1)) = n/2 (2 + 4n - 4) = n/2 (4n-2) = 2n^2-n ✅
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