Question

If S_n denotes the sum of the first n terms of an A.P., prove that S₃₀ = 3 (S₂₀ − S₁₀) ? 

Answer 1

We know that the sum of the n^th term is given by:

\[S_n = \frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right]\]

Here:
a = first term
d = common difference
n = number of terms in an A.P.
We need to prove the following:

\[S_{30} = 3\left( S_{20} - S_{10} \right)\]

Considering R.H.S., we get:

\[3\left( S_{20} - S_{10} \right)\]
\[ = 3\left[ \frac{20}{2}\left\{ 2a + \left( 20 - 1 \right)d \right\} - \frac{10}{2}\left\{ 2a + \left( 10 - 1 \right)d \right\} \right]\]

\[= 3\left[ 10\left( 2a + 19d \right) - 5\left( 2a + 9d \right) \right]\]

\[= 3\left[ 20a + 190d - 10a - 45d \right]\]

\[= 3\left[ 10a + 145d \right]\]

\[= 15\left[ 2a + 29d \right]\]

Considering L.H.S., we get:

\[S_{30} = \frac{30}{2}\left[ 2a + \left( 30 - 1 \right)d \right]\]
\[ = 15\left[ 2a + 29d \right]\]

L.H.S. = R.H.S.
Hence, proved.