Question

The $12th$ term of an AP is $-13$ and the sum of its first four terms is $24$. Find the sum of its first $10$ terms.

Answer 1

Step $1:$ Consider the first given condition and make an equation from that condition.

Assume that, $a$ be the first term of the given AP and $d$ be the common difference of the given AP.

Since the $12th$ term of the given AP is $-13$.

$$\begin{gathered} a_{12}=-13 \\ \Rightarrow a+11d=-13\cdots \left(1\right) \end{gathered}$$

Step $2:$ Consider the second given condition and make another equation from that condition.

Since the sum of its first four terms is $24$.

$$\begin{gathered} S_4=24 \\ \Rightarrow \frac{4}{2}\left\{2a+\left(4-1\right)d\right\}=24 \\ \Rightarrow 2\left\{2a+3d\right\}=24 \\ \Rightarrow 2a+3d=12\dots \left(2\right) \end{gathered}$$

Step $3:$ Solve equation $1$ and $2$ for $a,d$.

Multiply $2$ on both sides of equation $1$.

$2a+22d=-26$

So, equation $1$ becomes $2a+22d=-26$.

Now, subtract equation $2$ from equation $1$ and solve for $d$.

$$\begin{gathered} 2a+22d-2a-3d=-26-12 \\ \Rightarrow 19d=-38 \end{gathered}$$

Therefore, $d=-2$

Now, substitute $d=-2$ in equation $2$ and solve for $a$.

$$\begin{gathered} 2a+3\left(-2\right)=12 \\ \Rightarrow 2a-6=12 \\ \Rightarrow 2a=18 \end{gathered}$$

Therefore, $a=9$.

Step $4:$ Compute the required sum.

Substitute the value of $a,d$ and $n=10$ in the formula of the sum of AP and Simplify.

Sum of first $10$ terms $=\frac{10}{2}\left\{2\times 9+\left(10-1\right)\left(-2\right)\right\}$

$\Rightarrow$Sum of first $10$ terms $=5\left\{18-18\right\}$

Therefore, the sum of the first $10$ terms of the given AP is zero.