Question

Find the $31^{st}$ term of an AP whose $11^{th}$ term is $38$ and the $6^{th}$ term is $73$.

Answer 1

In this question, given that in AP the $11^{th}$ term is $38$ and the $6^{th}$ term is $73$. Here we need to find the $31^{st}$ term of the AP. AP stands for Arithmetic Progression. It is a sequence where the difference between the consecutive terms are the same. By using the formula of arithmetic progression, we can easily find the $31^{st}$ term of the AP.
Formula used :
The arithmetic sequence can be written as
$\\{ a,\ a + d,\ a + 2d,\ a + 3d,\ \ldots\ \\}$.
Formula used to find the $n^{th}$ terms in arithmetic progression is
$a_{n}= \ a\ + \ (n-1)\ \times \ d$
Where $a$ is the first term
$d$ is the common difference
$n$ is the number of term
${a_n}$ is the $n^{th}$ term

Complete step by step answer:
Given,
$11^{th}$ term is $38$
⇒ $a_{11}= 38$
By applying the formula,
We get,
$a + \left( 11 – 1 \right) \times d = 38$
By simplifying
We get,
$a + 10d = 38$
Let us consider this as equation (1),
$a = 38 – 10d$ ••• (1)
Also given,
$6^{th}$ term is $73$
⇒ $a_{6}= 73$
By applying the formula,
We get,
$a + \left( 6 – 1 \right) \times d = 73$
By simplifying,
We get,
$a + 5d = 73$
Let us consider this as equation (2),
$a = 73 – 5d$ •••(2)
By equating both (1) and (2) ,
We get,
$38 – 10d = 73 – 5d$
By bring all the variable terms to left side and the constants to right side,
We get,
$5d – 10d = 73 – 38$
By simplifying,
We get,
$- 5d = 35$
⇒ $d = - \dfrac{35}{5}$
By dividing,
We get,
$d = - 7$
Substituting the value of $d$ in (1)
(1)⇒ $a = 38 – 10d$
By substituting,
We get,
$a = 38 – 10\left( - 7 \right)$
By simplifying,
We get,
$a = 108$
Now we can find the $31^{st}$ term of an AP,
$a_{31}= a + \left( 31 – 1 \right) \times d$
By substituting the values,
We get,
$a_{31} = 108 + \left( 30 \times \left( - 7 \right) \right)\$
By simplifying,
We get,
$a_{31}= 108 – 210$
By subtracting,
We get,
$a_{31}= - 102$
Thus we get the $31^{st}$ term of an AP is $- 102$.

Note:
A simple example for arithmetic progression is the sequence of $2,6,10,14,\ldots$ here the first term $a$ is $2$ and the common difference $d$ is $4$ . The number $4$ is added to get the next terms. Similarly geometric progression (GP) is defined as a sequence where the ratio between the consecutive terms are the same. The formula to find the $n^{th}$ term is $a_{n}= \ ar^{n-1}$