Question

in an arithmetic progression. if tn=2n-1.then what is Sn?

Answer 1

Tn = 2n - 1

let n be 1

T1 = 2 - 1 = 1

Let n be 2

T2 = 4 - 1 = 3

Common difference ( d ) Will be 2


$sn = \frac{n}{2} (2a + (n - 1)d)$

$sn = \frac{n}{2} (2 + (n - 1) \times 2)$

$sn = \frac{n}{2} (2 + 2n - 2)$


$sn = \frac{n}{2} (2n)$


$sn = \frac{2 {n}^{2} }{2}$

Sn = n²

Answer 2

t1 will be 1
t2 will be 3
t3 will be 5 and so on.
hence d will be 3-1 =2
so
sn =
$\frac{n}{2} (2a + (n - 1)d) \\ \frac{n}{2} (2 + (n - 1)2) \\ {n}^{2}$