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1. [4 marks]
In an arithmetic sequence, the sum of the 3rd and 8th terms is 1.
Given that the sum of the first seven terms is 35, determine the first term and the common difference.​

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LammettHash

An arithmetic sequence [tex]a_n[/tex] has a fixed difference [tex]d[/tex] between consecutive terms, so that they are recursively described by

[tex]a_n=a_{n-1}+d[/tex]

We're told that the sum of the 3rd and 8th terms is 1, so

[tex]a_3+a_8=1[/tex]

Using the recursive rule above, we have

[tex]a_8=a_7+d[/tex]

[tex]a_8=(a_6+d)+d=a_6+2d[/tex]

[tex]a_8=(a_5+d)+2d=a_5+3d[/tex]

and so on down to

[tex]a_8=a_3+5d[/tex]

which means

[tex]a_3+(a_3+5d)=\boxed{2a_3+5d=1}[/tex]

More generally, we can do the same manipulation with the recursive rule to find the explicit rule:

[tex]a_n=(a_{n-2}+d)+d=a_{n-2}+2d[/tex]

[tex]a_n=(a_{n-3}+d)+2d=a_{n-3}+3d[/tex]

and so on down to

[tex]a_n=a_3+(n-3)d[/tex]

We're also told that the sum of the first 7 terms is 35:

[tex]a_1+a_2+a_3+a_4+a_5+a_6+a_7=35[/tex]

and using the explicit rule above, this is the same as

[tex](a_3-2d)+(a_3-d)+a_3+(a_3+d)+(a_3+2d)+(a_3+3d)+(a_3+4d)=35[/tex]

[tex]\implies7a_3+7d=35\implies\boxed{a_3+d=5}[/tex]

Now solve for [tex]a_3[/tex] and [tex]d[/tex]:

[tex]a_3+d=5\implies d=5-a_3[/tex]

[tex]2a_3+5d=1\implies2a_3+5(5-a_3)=1[/tex]

[tex]\implies25-3a_3=1[/tex]

[tex]\implies3a_3=24[/tex]

[tex]\implies a_3=8[/tex]

The common difference is then

[tex]d=5-8\implies\boxed{d=-3}[/tex]

and in turn, the first term is

[tex]a_1=a_3-2d=8-2(-3)\implies\boxed{a_1=14}[/tex]

The first term of the arithmetic sequence is 14, while the common difference is -3

The process of calculating the above values are presented as follows;

The known parameters are;

The type of sequence = Arithmetic sequence

The sum of the 3rd term, t₃, and the 8th term, t₈ = 1

The sum of the first  seven terms of the sequence, S₇ = 35

Unknown;

The first term, a

The common difference, d

Method;

The nth term of an AP, tₙ is tₙ = a + (n - 1)·d

The sum of n terms of an arithmetic progression, Sₙ, is given as follows;

[tex]\mathbf{S_n = \dfrac{n}{2} \left[2 \cdot a + \left(n - 1\right) \cdot d \right ] = \dfrac{n}{2} \left[a + l ]}[/tex]

Plugging in the given values gives;

t₃ = a + (3 - 1)·d = a + 2·d

t₈ = a + (8 - 1)·d = a + 7·d

t₃ + t₈ = a + 2·d + a + 7·d = 1

∴ 2·a + 9·d = 1...(1)

[tex]\mathbf{S_7 = 35 = \dfrac{7}{2} \left[2 \cdot a + \left(7 - 1\right) \cdot d \right ] = 7 \cdot a+ 21 \cdot d}[/tex]

Therefore;

7·a + 21·d = 35...(2)

Making a the subject of both equation (1) and (2) gives;

From equation (1), a = (1 - 9·d)2 = 0.5 - 4.5·d

From equation (2), a = (35 - 21·d)/7 = 5 - 3·d

Equating both values of a gives;

∴ 0.5 - 4.5·d = 5 - 3·d

0.5 - 5 = 4.5·d - 3·d = 1.5·d

∴ d = -4.5×1.5 = -3

The common difference of the arithmetic progression, d = -3

a = 0.5 - 4.5·d

∴ a = 0.5 - 4.5 × (-3) = 14

The first term of the arithmetic progression, a = 14

Learn more about arithmetic progression here;

https://brainly.com/question/7448329

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