# Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show thatS3 = 3(S2 – S1)

Let a and b be the first term and the common difference of the A.P. respectively.

Therefore,

S1 = (n/2)[2a + (n - 1)d] …(1)

S2 = (2n/2)[2a + (2n - 1)d] …(2)

S3 = (3n/2)[2a + (3n - 1)d] …(3)

From (1) and (2), we obtain

S2 - S1 = (2n/2)[2a + (2n - 1)d] - (n/2)[2a + (n - 1)d]

= (n/2)[{4a + (4n - 2)d} - {2a + (n - 1)d}]

= (n/2)[4a + 4nd - 2d - 2a - nd + d]

= (n/2)[2a + 3nd - d]

= (1/3) × (3n/2)[2a + (3n - 1)d]

= (1/3)S3

Thus, S3 = 3(S2 - S1)

Hence, the given result is proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Interactive quiz on inequalities involving AM and GM45 mins  Check Your progress Part 2| Interactive Quiz: Sequence & Series47 mins  Sequence & Series (Lecture 1)56 mins  Sequence & Series (Lecture 5)42 mins  Sequence & Series (Lecture 7)37 mins  Trigonometric Series45 mins  Quick Recap lecture of Sequence & Series56 mins  General Term of Miscellaneous progression46 mins  Improve your understanding of AP & GP27 mins  Geometric Progression58 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 