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).

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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.

S

_{n}= n[2a +(n – 1)d] / 2We get

S

_{30}= 30[2a + (30 – 1)d] / 2= 15[2a + 29d]

R.H.S. = 3(S

_{20}– S_{10}).= 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]

= S

_{30}Hence proved.