Given:
The 78 terms are 655
The 5 terms are 71
Find-:
The value of 39 terms
Explanation-:
The terms value is
[tex]\begin{gathered} a_{78}=655 \\ \\ a_5=71 \end{gathered}[/tex]The nth term of an arithmetic sequence is:
[tex]a_n=a+(n-1)d[/tex]Where,
[tex]\begin{gathered} a_n=n^{th}\text{ terms of the sequence} \\ \\ a=\text{ First term of the sequence} \\ \\ d=\text{ Common difference} \end{gathered}[/tex]The 78 terms are
[tex]\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}[/tex]The 5 terms are
[tex]\begin{gathered} a_n=a+(n-1)d \\ \\ n=5 \\ \\ a_5=71 \\ \\ 71=a+(5-1)d \\ \\ 71=a+4d...........(2) \end{gathered}[/tex]The eq(2) - eq(1) is
[tex]\begin{gathered} 655-71=a+77d-(a+4d) \\ \\ 584=a+77d-a-4d \\ \\ 584=73d \\ \\ d=\frac{584}{73} \\ \\ d=8 \end{gathered}[/tex]The value of "d" is 8.
The value of "a" is
[tex]\begin{gathered} 71=a+4d \\ \\ 71=a+4(8) \\ \\ 71=a+32 \\ \\ a=71-32 \\ \\ a=39 \end{gathered}[/tex]The 39 terms are:
[tex]\begin{gathered} a_n=a+(n-1)d \\ \\ a=39 \\ \\ d=8 \\ \\ n=39 \end{gathered}[/tex]So, the 39 terms
[tex]\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}[/tex]The value of 39 terms is 343