Answer :

From the question we know that, Sn = n2 + 2n ……. (1)


We know that, Sn = × (2a + (n – 1)d)


= na +


= na + (n2 – n) ×


Sn = na + n2 × – n × ……. (2)


As we know that, (1) and (2) are both sum of the same arithmetic progression


So we can equate them to each other.


So, we have,


n2 + 2n = na + n2 × – n ×


n2 + 2n = n2 × + na – n ×


Now we will equate the coefficients of “n2” and “n” on both the sides of the equal to sign.


So, we get,


For coefficients of n2 :


1 =


d=2


Now, For coefficients of n:


2 = a –


From above we have that d=2,


So we can say that,


2 = a –


2 = a – 1


a=3


So, now, we know that, Tn = a + (n – 1)d


Now we put values of a and n in the above equation,


We get,


Tn = 3 + (n – 1)2


Tn = 3 + 2n – 2


Tn = 2n + 1


for the given value of Sn = n2 + 2n, we have Tn = 2n + 1.


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