If $S_n$ denotes the sum of the first $n$ terms of an A.P., prove that $S_{30} = 3 (S_{20} – S_{10})$.

Given:

$S_n$ denotes the sum of first $n$ terms of an A.P.

To do:

We have to prove that $S_{30} = 3(S_{20} – S_{10})$.

Solution:

Let $a$ be the first term and $d$ the common difference.

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{10}=\frac{10}{2}[2 \times a+(10-1) \times d]$

$=5[2a+9d]$

$=10a+45d$......(i)

$S_{20}=\frac{20}{2}[2 \times a+(20-1) \times d]$

$=10[2a+19d]$

$=20a+190d$......(ii)

$S_{30}=\frac{30}{2}[2 \times a+(30-1) \times d]$

$=15[2a+29d]$

$=30a+435d$......(iii)

From (i) and (ii)

$3(S_{20}-S_{10})=3[20a+190d-(10a+45d)]$

$=3(20a+190d-10a-45d)$

$=3(10a+145d)$

$=30a+435d$

$=S_{30}$       (From (iii))

Hence proved.