Question

The nth term of an Arithmetic Progression (A.P.) is given by the relation T_n = 6(7 − n). Find:

1. its first term and the common difference
2. sum of its first 25 terms

Answer 1

(i) The first term a or T₁, substitute n = 1 into the given relation:

T₁ = 6(7 − 1)

T₁ = 6 × 6

T₁ = 36

The second term (T₂) by substituting n = 2:

T₂ = 6(7 − 2)

T₂ = 6 × 5

T₂ = 30

d = T₂ − T₁

d = 30 − 36

d = −6

(ii) `S_n = n/2[2a + (n - 1)d]`

`S_25 = 25/2[2(36) + (25 - 1)(-6)]`

= 12.5[72 + 24 × (−6)]

= 12.5[72 − 144)]

= 12.5 × (−72)

= −900