Find the sum of t

(i)to 10 terms

S_n = [2a + (n – 1) d]

S₁₀ = [100 + 9(-4)]

= 5 [100 – 36]

= 5(64) = 320

(ii)to 12 terms

S_n = [2a + (n – 1) d]

= [2 + (12 – 1) 2]

= 6 (2 + 22)

= 144

(iii)to 25 terms

S_n = [2a + (n – 1) d]

= [2 (3) + 24()]

= [6 + 36]

= 525

(iv)to 12 terms

S_n = [2a + (n – 1) d]

= [82 - 55]

= 6 [27] = 162

(v)to 22 terms

S_n = [2a + (n – 1) d]

= [2(a + b) + (21) (-2b)]

= 11 [2a + 2b – 42b]

= 11 [2a – 40b]

= 22a – 440b

(vi)to n terms

S_n = [2a + (n – 1) d]

= [2(x – y)² + (n – 1) (2xy)]

= (2) [(x – y)² + (n – 1) xy]

= n [(x – y)² + (n – 1) xy]

(vii)to n terms

S_n = [2a + (n – 1) d]

= [2() + (n – 1) xy]

= {2(x – y) + (n – 1) (2x – y)}

= {n (2x – y) – y}

(viii)to 36 terms

S_n = [2a + (n – 1) d]

= [-52 + (35) 2]

= 18 [-52 + 70]

= 18 [18] = 324