Question

Question 5 of 15For a given arithmetic sequence, the 78th term, a7g, is equal to 655, and the 5th term, as, is equal to71.Find the value of the 39th term, a3913th term

Answer 1

Given:

The 78 terms are 655

The 5 terms are 71

Find-:

The value of 39 terms

Explanation-:

The terms value is

`\begin{gathered} a_(78)=655 \\ \\ a_5=71 \end{gathered}`

The nth term of an arithmetic sequence is:

`a_n=a+(n-1)d`

Where,

`\begin{gathered} a_n=n^(th)\text{ terms of the sequence} \\ \\ a=\text{ First term of the sequence} \\ \\ d=\text{ Common difference} \end{gathered}`

The 78 terms are

`\begin{gathered} a_n=a+(n-1)d \\ \\ n=78 \\ \\ a_(78)=655 \\ \\ a_(78)=a+(78-1)d \\ \\ 655=a+77d...........(1) \end{gathered}`

The 5 terms are

`\begin{gathered} a_n=a+(n-1)d \\ \\ n=5 \\ \\ a_5=71 \\ \\ 71=a+(5-1)d \\ \\ 71=a+4d...........(2) \end{gathered}`

The eq(2) - eq(1) is

`\begin{gathered} 655-71=a+77d-(a+4d) \\ \\ 584=a+77d-a-4d \\ \\ 584=73d \\ \\ d=(584)/(73) \\ \\ d=8 \end{gathered}`

The value of "d" is 8.

The value of "a" is

`\begin{gathered} 71=a+4d \\ \\ 71=a+4(8) \\ \\ 71=a+32 \\ \\ a=71-32 \\ \\ a=39 \end{gathered}`

The 39 terms are:

`\begin{gathered} a_n=a+(n-1)d \\ \\ a=39 \\ \\ d=8 \\ \\ n=39 \end{gathered}`

So, the 39 terms

`\begin{gathered} a_n=a+(n-1)d \\ \\ a_(39)=39+(39-1)8 \\ \\ a_(39)=39+38(8) \\ \\ a_(39)=39+304 \\ \\ a_(39)=343 \end{gathered}`

The value of 39 terms is 343