Question

The sum of the first 9 terms of an arithmetic sequence is 45 and the sum of first 18 terms is 171.
(a) What is the sum of its $${10}^{th}$$ to $${18}^{th}$$ terms?
(b) What is its $$5^{th}$$ term?
(c) Find its $${14}^{th}$$ term.
(d) Find the sum of $$5^{th}$$ to $${14}^{th}$$ terms.

Answer 1

Hint:
Here, we will substitute the given values in the formula of the sum of $$n$$ terms of an AP to find two equations. Then we will use the elimination method to get the first term and common difference of the given AP. Using these values we will be able to find the answers to all the given parts.

Formula Used:
We will use the following formulas:
1. The general term of an AP is $$a_n=a+\left(n-1\right)d$$ where, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the total number of terms in an AP.
2. The sum of $$n$$ terms of an AP is given by $$S_n={\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]$$

Complete step by step solution:
Let the first term of an Arithmetic Progression (AP) be $$a$$, the common difference be $$d$$ and the total number of terms in the AP be $$n$$
The sum of first 9 terms of an arithmetic sequence is 45.
Hence, in this situation, $$n=9$$ and $$S_9=45$$
Thus, substituting these values in the formula of sum of $$n$$ terms $$S_n={\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]$$, we get
$$S_9={\displaystyle \frac{9}{2}}\left[2a+\left(9-1\right)d\right]$$
$$\Rightarrow 45={\displaystyle \frac{9}{2}}\left[2a+8d\right]$$
Taking 2 common from the bracket and cancelling it out with the denominator, we get
$$\Rightarrow 45=9\left[a+4d\right]$$
Dividing both sides by 9, we get
$$\Rightarrow a+4d=5$$……………………………..$$\left(1\right)$$
Similarly, it is given that the sum of the first 18 terms is 171.
Thus, substituting $$n=18$$ and $$S_{18}=171$$ in the formula of sum of $$n$$ terms $$S_n={\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]$$, we get
$$S_{18}={\displaystyle \frac{18}{2}}\left[2a+\left(18-1\right)d\right]$$
$$\Rightarrow 171=9\left[2a+17d\right]$$
Dividing both sides by 9, we get
$$\Rightarrow 2a+17d=19$$……………………………$$\left(2\right)$$
Now, we will solve the equations $$\left(1\right)$$ and $$\left(2\right)$$ using elimination method.
We will multiply equation $$\left(1\right)$$ by 2 and then subtract the second equation from it.
Thus, we get,
$$\Rightarrow 2a+8d-2a-17d=10-19$$
Adding and subtracting the like terms, we get
$$\Rightarrow -9d=-9$$
Dividing both sides by $$-9$$, we get
$$\Rightarrow d=1$$
Hence, substituting this in equation $$\left(1\right)$$, we get
$$a+4\left(1\right)=5$$
$$\Rightarrow a=5-4=1$$
Therefore, the first term of the AP, $$a=1$$ and the common difference, $$d=1$$
Now,
(a) What is the sum of its $${10}^{th}$$ to $${18}^{th}$$ terms?
First of all, we will find the $${10}^{th}$$ term
Hence, $$a_{10}=a+\left(10-1\right)d$$
Substituting the values of the first term and common difference, we get,
$$a_{10}=1+9\times 1=1+9=10$$
Hence, to find the sum from $${10}^{th}$$ to $${18}^{th}$$ terms of this AP, we will consider the first term as the $${10}^{th}$$ term that is, $$a=10$$ for this case.
Also, from the $${10}^{th}$$ to $${18}^{th}$$ terms there are total $$18-10+1=9$$ terms
Hence, we will take $$n=9$$ for this case.
Thus sum of these 9 terms from $${10}^{th}$$ to $${18}^{th}$$ terms of this AP will be:
$$S_9={\displaystyle \frac{9}{2}}\left[2\left(10\right)+\left(9-1\right)\left(1\right)\right]$$, where the common difference will remain the same
$$\Rightarrow S_9={\displaystyle \frac{9}{2}}\left[20+8\right]={\displaystyle \frac{9}{2}}\times 28=9\times 14=126$$
Therefore, the sum of $${10}^{th}$$ to $${18}^{th}$$ terms of this AP is 126.
(b) What is its $$5^{th}$$ term?
Now, in order to find the $$5^{th}$$ term of this AP, we will use the general formula, i.e.
$$a_n=a+\left(n-1\right)d$$
Here, substituting $$n=5$$, $$a=1$$ and $$d=1$$, we get,
$$\Rightarrow a_5=1+\left(5-1\right)\left(1\right)$$
$$\Rightarrow a_5=1+4=5$$
Therefore, the fifth term of this AP is 5.
(c) Find its $${14}^{th}$$ term.
Similarly, like the (b) part, we will repeat the steps but in this case $$n=14$$
$$\Rightarrow a_{14}=1+\left(14-1\right)\left(1\right)$$
$$\Rightarrow a_{14}=1+13=14$$
Therefore, the $${14}^{th}$$ term of this AP is 14.
(d) Find the sum of $$5^{th}$$ to $${14}^{th}$$ terms.
In this part, we will again repeat the steps of the (a) part.
Hence, in this case, we will consider the first term as $$a=5$$ because we know that the fifth term is 5.
And the total number of terms is $$14-5+1=10$$
Hence, $$S_{10}={\displaystyle \frac{10}{2}}\left[2\left(5\right)+\left(10-1\right)\left(1\right)\right]$$
$$\Rightarrow S_{10}=5\left[10+9\right]=5\times 19=95$$

Therefore, the sum of $$5^{th}$$ to $${14}^{th}$$ terms of this AP is 95.
Thus, this is the required answer.

Note:
An Arithmetic Progression is a sequence of numbers such that the difference between any term and its preceding term is constant. This difference is known as the common difference of an Arithmetic Progression (AP). A real life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.