Hey!! how y'all doing? I need the answer of this question. (21 pts each)
Answers
Kindly refer to the attachment for solution ^_^
Hope it helps ^_^
Answer:
S₁ + S₃ = 2S₂
Step-by-step explanation:
Sum of n terms of an AP sn = (n/2)[2a + (n - 1) * d]
(i) S₁
First term(a) = 1
Common difference(d) = 1
Sum of n terms = (n/2)[2 + (n - 1) * 1]
= (n/2)[2 + n - 1]
= (n/2)[n + 1]
= n(n+1)/2
(ii) S₂:
First term(a) = 1
Common difference(d) = 2
Sum of n terms = (n/2)[2a + (n - 1) * d]
= (n/2)[2 + (n - 1) * 2]
= (n/2)[2 + 2n - 2]
= (n/2)[2n]
= n²
(iii) S₃:
First term(a) = 1
Common difference (d) = 3
Sum of n terms = (n/2)[2a + (n - 1) * d]
= (n/2)[2 + (n - 1) * 3]
= (n/2)[2 + 3n - 3]
= (n/2)[3n - 1]
= n(3n - 1)/2
LHS:
S₁ + S₃ = {n(n+1)/2} + {n(3n - 1)/2}
= n[n + 1 + 3n - 1]/2
= n[n + 3n]/2
= n[4n]/2
= 2n²
= 2S₂ {From (ii)}
= RHS
Therefore, S₁ + S₃ = 2S₂
Hope it helps!