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xKelvin

Answer:

[tex]\displaystyle a_1 = 44[/tex]

Step-by-step explanation:

Recall the direct formula for an arithmetic sequence:
[tex]\displaystyle a_n = a_1 + d(n-1)[/tex]

Where d is the common difference.

Therefore, we can write the following two equations:
[tex]\displaystyle \left \{ {a_4 = 107 = a_1 + d(4-1)\atop {a_8 = 191 = a_1 + d(8-1)}} \right.[/tex]

Solve the system. Simplifying yields:
[tex]\displaystyle 107 = a_1 + 3d\text{ and } 191 = a_1 + 7d[/tex]

Subtracting the two equations into each other yields:
[tex]\displaystyle \begin{aligned} (191) - (107) & = (a_1 + 7d) - (a_1 + 3d) \\ \\ 84 & = 4d \\ \\ d & = 21 \end{aligned}[/tex]

Using either equation, solve for the initial term:

[tex]\displaystyle \begin{aligned} 107 & = a_1 + 3(21) \\ \\ a_1 & = 107-63 \\ \\ & = 44\end{aligned}[/tex]

In conclusion, the initial term is 44.

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Answer:

[tex]\huge\boxed{\bf\:a_{1} = 44}[/tex]

Step-by-step explanation:

In the given arithmetic series,

  • [tex]a_{4} = 107[/tex]
  • [tex]a_{8} = 191[/tex]
  • [tex]a_{1} = ?[/tex]

We know that,

[tex]\boxed{a_{n} = a + (n - 1)d}[/tex]

Therefore,

[tex]a_{4} = a + 3d = 107 -----(1)\\a_{8} = a + 7d = 191 -----(2)[/tex]

By solving the two equations (1) & (2)....

[tex]\:\:\:\:a + 7d = 191 \\-\underline{a + 3d = 107 }\\ \underline{\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:4d = 84\:\:\:\:\:\:\:\:}}\\\\\\\\4d = 84\\d = 84 \div 4\\\boxed{d = 21}[/tex]

Now, we have the value of the common difference (d). To find [tex]a_{1}[/tex] or a, let's substitute the value of 'd' in (1).

[tex]a + 3d =107\\a + 3(21)=107\\a + 63 = 107\\a = 107 - 63\\\boxed{\bf\:a_{1} = 44}[/tex]

[tex]\rule{150pt}{2pt}[/tex]

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