Question

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

Answer 1

Hint: 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 \[a_{n}= \ ar^{n-1}\]