Q. 25.0( 1 Vote )

# If Sn

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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