This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.6
In this question we are given that Sn denotes the sum of the first n terms of an A.P.,
Now we have to prove that S30 = 3(S20 – S10).
CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 70
Let’s take L.H.S.
By using the formula of the sum of n terms of an A.P.
Sn = n[2a +(n – 1)d] / 2
We get
S30 = 30[2a + (30 – 1)d] / 2
= 15[2a + 29d]
R.H.S. = 3(S20 – S10).
= 3[20[2a + (20 – 1)d] / 2 – 10[2a + (10 – 1)d] / 2]
= 3[10(2a + 19d) – 5(2a + 9d)]
= 3[20a + 190d – 10a – 45d]
= 3[10a + 145d]
= 3 × 5[2a + 29d]
= 15[2a + 29d]
= S30
Hence proved.