Given t_{19} = 52 and t_{38} = 128

Now we have to find the value of “a” and “d”

Using n^{th} term of an A.P. formula t_{n} = a + (n – 1) d

where n = no. of terms

a = first term

d = common difference

t_{n} = n^{th} terms

we will find value of “a” and “d”

Let, t_{19} = a + (19 – 1) d

⇒ 52 = a + 18 d …. (1)

t_{38} = a + (38 – 1) d

⇒ 128 = a + 37 d …. (2)

Subtracting equation (1) from equation (2), we get,

⇒ 128 – 52 = (a – a) + (37 d – 18 d)

⇒ 76 = 19 d

⇒ d = 76/19 = 4

Substitute value of “d” in equation (1) to get value of “a”

⇒ 52 = a + 18 ×4

⇒ 52 = a + 72

⇒ a = 52 – 72 = – 20

Now, to find value of S_{56} we will using formula of sum of n terms

Where, n = no. of terms

a = first term

d = common difference

S_{n} = sum of n terms

Thus, substituting given value in formula we can find the value of S_{n}

⇒S_{56} = 28 × [ – 40 + 55×4]

⇒S_{56} = 28 × [ – 40 + 220]

⇒S_{56} = 28 × 180 = 5040

Thus, S_{56} = 5040