24. If SI, S2 and S3 be the sum of n, 2n and 3n terms respectively of

1.	24. If SI, S2 and S3 be the sum of n, 2n and 3n terms respectively of an AP, show that83 3(S2 - s).
Answer» S₁=Sₙ=n/2 (2a+ (n-1) d) s₂= 2n/2 {2a+ (2n-1) d} s₃=3n/2 {2a+ (3n-1)d} solving RHS 3{s₂-s₁) 3 [2n/2 {2a+ (2n-1)d} - n/2 {2a+ (n-1)d }] 3n/2 [2{2a+(2n-1)}]-[2a+(n-1)d] 3n/2 [4a+2(2n-1)d-2a-(n-1)d] 3n/2 [4a+ 4nd- 2d- 2a- nd+ d] 3n/2 [2 a+ 3nd-d] 3n/2 [2a+ (3n-1) d]=s₃ Please like the solution

24. If SI, S2 and S3 be the sum of n, 2n and 3n terms respectively of an AP, show that83 3(S2 - s).

Answer»

S₁=Sₙ=n/2 (2a+ (n-1) d)

s₂= 2n/2 {2a+ (2n-1) d}

s₃=3n/2 {2a+ (3n-1)d}

solving RHS

3{s₂-s₁)

3 [2n/2 {2a+ (2n-1)d} - n/2 {2a+ (n-1)d }]

3n/2 [2{2a+(2n-1)}]-[2a+(n-1)d]

3n/2 [4a+2(2n-1)d-2a-(n-1)d]

3n/2 [4a+ 4nd- 2d- 2a- nd+ d]

3n/2 [2 a+ 3nd-d]

3n/2 [2a+ (3n-1) d]=s₃

Please like the solution