Question

If ${{S}_{n}}$ denotes the sum of the first $n$ terms of an A.P., prove that ${{S}_{30}}=3\left( {{S}_{20}}-{{S}_{10}} \right)$

Answer 1

Hint: In this question we have been told that ${{S}_{n}}$ denotes the sum of the first $n$ terms of an A.P. the A.P referred to here is an arithmetic progression. We will consider the first term of the arithmetic progression to be $a$, the common difference be $d$ and $n$ be the number of terms. we will use the formula of the sum of $n$ terms of an A.P which is given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . we will then use this formula to find ${{S}_{30}},{{S}_{20}}$ and ${{S}_{10}}$, and equate to get the required solution.

Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow {{S}_{30}}=3\left( {{S}_{20}}-{{S}_{10}} \right)\to \left( 1 \right)$
Now using the formula of the sum of $n$ terms, we get ${{S}_{30}}$ as:
$\Rightarrow {{S}_{30}}=\dfrac{30}{2}\left[ 2a+\left( 30-1 \right)d \right]\to \left( 2 \right)$
Now using the formula of the sum of $n$ terms, we get ${{S}_{20}}$ as:
$\Rightarrow {{S}_{20}}=\dfrac{20}{2}\left[ 2a+\left( 20-1 \right)d \right]$
Now using the formula of the sum of $n$ terms, we get ${{S}_{30}}$ as:
$\Rightarrow {{S}_{10}}=\dfrac{10}{2}\left[ 2a+\left( 10-1 \right)d \right]$
Now consider the right-hand side of equation $\left( 1 \right)$, we have:
$\Rightarrow 3\left( {{S}_{20}}-{{S}_{10}} \right)$
On substituting the value of ${{S}_{20}}$ and ${{S}_{10}}$, we get:
$\Rightarrow 3\left( \dfrac{20}{2}\left[ 2a+\left( 20-1 \right)d \right]-\dfrac{10}{2}\left[ 2a+\left( 10-1 \right)d \right] \right)$
On simplifying, we get:
$\Rightarrow 3\left( 10\left[ 2a+19d \right]-5\left[ 2a+9d \right] \right)$
On multiplying the terms, we get:
$\Rightarrow 3\left( 20a+190d-10a-45d \right)$
On simplifying, we get:
$\Rightarrow 3\left( 10a+145d \right)$
On taking the term $5$ common from the terms, we get:
$\Rightarrow 15\left( 2a+29d \right)$
Now we can write $15$ as $\dfrac{30}{2}$ therefore, on substituting, we get:
$\Rightarrow \dfrac{30}{2}\left( 2a+29d \right)$
Now we can write $29$ as $30-1$ therefore, on substituting, we get:
$\Rightarrow \dfrac{30}{2}\left( 2a+\left( 30-1 \right)d \right)$
We can see the expression is the right-hand side of the expression therefore, we get the value of the above expression as:
$\Rightarrow {{S}_{30}}$, which is the left-hand side of equation $\left( 1 \right)$, hence proved.

Note: In this question we have used the formula for the sum of $n$ terms in an arithmetic progression. the general formula for the ${{n}^{th}}$ term of an arithmetic progression should be remembered which is given by ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$, where ${{a}_{n}}$ is the ${{n}^{th}}$ term in the sequence, ${{a}_{1}}$ is the first term of the sequence and $d$ is the common difference between any two terms.