Given :

a₃ = a + d =

⇒ a = – d ……… (1)

S₁₀ = but take S₁₀ as 175

i.e., S₁₀ =

We know that,

⇒ = [a+d]

⇒ = ( – d) + d [∵ a = 15 – 2d]

⇒ = – d + d

⇒ – = d

Substituting d = in equation (1) we get

a = – 2 × = – =

Now, a_n = a + (n – 1) d

a₁₀ = a + d = + × 1 = + =

∴ a₁₀ = ; d =