Question

In an A.P. t₁₀ = 57 and t₁₅ = 87, then find t₂₁ 

Answer 1

For an A.P., let a be the first term and d be the common difference.

t₁₀ = 57 and t₁₅ = 87     ......[Given]

Since t_n = a + (n – 1)d

t₁₀ = a + (10 – 1)d

∴ 57 = a + 9d

i.e., a + 9d = 57     ......(i)

Also, t₁₅ = 87

∴ a + (15 – 1)d = 87

∴ a + 14d = 87   ......(ii)

Subtracting equation (i) from (ii), we get

a + 14d = 87
a +  9d = 57
–      –       –   
       5d = 30

∴ d = `30/5` = 6

Substituting d = 6 in equation (i), we get

a + 9(6) = 57

∴ a + 54 = 57

∴ a = 57 – 54 = 3

t₂₁ = a + (21 – 1)d

= a + 20d

= 3 + 20(6)

= 3 + 120

∴ t₂₁ = 123