For a given A.P. with

S₃ = 9, S₇ = 49, find S_n and S₁₀.

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

⇒ S₃ = 9,

S₃ = [2a + (3–1)d]

9 = [2a + 2d]

9 = [2(a + d)]

= [a + d]

3 = [a + d]

⇒ a = 3–d ……eq 1

⇒ S₇ = 49

S₇ = [2a + (7–1)d]

49 = [2a + 6d]

49 = [2(a + 3d)]

= a + 3d

7 = a + 3d

⇒ a = 7–3d ......eq 2

By equating eq 1 and eq 2

3–d = 7–3d

3d – d = 7 – 3

2d = 4

d = = 2

Putting value of d in eq 1

a = 3–d = 3–2

= 1

S_n = [2a + (n–1)d]

= [2×1 + (n–1)2]

= [2 + 2n–2]

= ×2n

= n²

∴ S_n = n²

⇒ S₁₀ = (10)²

= 100