If S1, S2, S3 denote respectively the sum of first n1, n2 and n3 terms of an A.P., then S1/n1 (n2-n3) +S2/n2 (n3-n1) + S3/n3(n1 -n2)

Question

If S₁, S₂, S₃ denote respectively the sum of first n₁, n₂ and n₃ terms of an A.P., then  \(\frac{S_1}{n_1}\)(n₂ -n₃) +\(\frac{S_2}{n_2}\)(n₃ -n₁) +\(\frac{S_3}{n_3}\)(n₁ -n₂) is equal to 

a) 0 

(b) n₁ + n₂ + n₃ 

(c) n₁ n₂ n₃ 

(d) S₁ S₂ S₃

Answer 1

Answer : (a) 0

Let the first term of the given A.P. be a and the common difference be d.

Then,

\( S_1 = \frac{n_1}{2}[2a+(n_1-1)d]\) 

\( S_2 = \frac{n_2}{2}[2a+(n_2-1)d]\)

\( S_3 = \frac{n_3}{2}[2a+(n_3-1)d]\)

= \(\frac{1}{2}\) [2a (n₂ – n₃) + n₁ (n₂ – n₃) – d(n₂ – n₃)]  

+ \(\frac{1}{2}\)[2a (n₃ – n₁) + n₂ (n₃ – n₁) – d (n₃ – n₁)] + \(\frac{1}{2}\) [2a (n₁ – n₂) + n₃ (n₁ – n₂) – d (n₁ – n₂)]

 \(\frac{1}{2}\) [2an₂ – 2an₃ + 2an₃ – 2an₁ + 2an₁ – 2an₂) + (n₁n₂ – n₁ n₃ + n₂n₃ – n₁n₂ + n₃n₁ – n₃n₂) – d (n₂ – n₃ + n₃ – n₁ + n₁ – n₂)] 

= 0.