Question

Which term of the arithmetic progression 5, 15, 25 . . . . . . will be 130 than its ${{31}^{st}}$ term?

Answer 1

Hint: Here, we will use basic formula of arithmetic progression of finding ${{n}^{th}}$ term which is ${{a}_{n}}=a+\left( n-1 \right)d$
Where ‘a’ is the first term of given AP, d is a common difference, and n is the number at which term is to be found.

Complete step by step answer:
As the given arithmetic progression is 5, 15, 25 . . . . . . , we can clearly see its first term is 5. So, a = 5
Subtracting the second term from the first term to obtain common difference we get 15-5 = 10, therefore, the common difference is 10. So, we can write d = 10.
By using formula of finding ${{n}^{th}}$ term in a given arithmetic progression ${{a}_{n}}=a+\left( n-1 \right)d$ we can calculate ${{a}_{31}}$
Putting n = 31 in ${{a}_{n}}=a+\left( n-1 \right)d$ we get:
\[{{a}_{31}}=a+\left( 31-1 \right)d\]
Putting values of a and d, we get:
\[\begin{align}
  & {{a}_{31}}=5+\left( 30 \right)10 \\
 & \Rightarrow {{a}_{31}}=5+300 \\
 & \Rightarrow {{a}_{31}}=305\cdots \cdots \cdots \left( i \right) \\
\end{align}\]
Hence, ${{31}^{st}}$ term is 305 which will be used to find required term.
Let the required term be ${{n}^{th}}$ term,
As in question, we have to find term which is 130 more than ${{31}^{st}}$ term we can write it as ${{a}_{n}}=130+{{a}_{31}}$
Using ${{a}_{n}}=a+\left( n-1 \right)d$ and putting value of ${{a}_{31}}$ from (i) we get:
\[a+\left( n-1 \right)d=130+305\]
Also, as calculated before, a = 5 and d = 10. Putting them in above equation, we get:
\[\begin{align}
  & 5+\left( n-1 \right)10=130+305 \\
 & \Rightarrow 5+10n-10=435 \\
 & \Rightarrow 10n=435+5 \\
 & \Rightarrow n=\dfrac{440}{10} \\
 & \Rightarrow n=44 \\
\end{align}\]
Hence, ${{44}^{th}}$ term is the required term. But we can also find the value of n = 44 by putting it in ${{a}_{n}}=a+\left( n-1 \right)d$ we get:
\[\begin{align}
  & {{a}_{44}}=5+\left( 44-1 \right)10 \\
 & \Rightarrow {{a}_{44}}=5+\left( 43 \right)\left( 10 \right) \\
 & \Rightarrow {{a}_{44}}=5+430 \\
 & \Rightarrow {{a}_{44}}=435 \\
\end{align}\]
Hence, ${{44}^{th}}$ term is 435 which is the required term.

Note:
Students should not get confused with $n\text{ and }{{a}_{n}}$. Here, n is the number at which value is required, and ${{a}_{n}}$ is a value at ${{n}^{th}}$ place is given arithmetic progression. Students should carefully find the common difference. It should always be subtracting the first term from the second term. Hence, the common difference can be negative too.