Sequences: past paper questions & worked solutions

The common difference is \(8-14=-6\).

Each successive term decreases by 6.

Use first term plus (n-1) common differences.

The common difference is \(-7\).

The common difference is \(-6\).

The sequence increases by \(4\), so

For the sequence \(5n+4\), take \(n=3\):

and \(19\) is prime.

The common difference is \(3\).

For the other sequence, \[ 5n+17 \]

The common difference is \(4\), so the nth term is \[ 4n-3 \]

For the second sequence, the nth term is \(3n+5\).

The \((2m)\)th term is \[ 3(2m)+5=6m+5 \]

Mario saves from 2021 up to and including 2070.

Number of years:

This is an arithmetic sequence.

First term:

Common difference:

Last term:

Use the arithmetic series formula:

For an arithmetic series,

and

So the sequence is

with common difference \(7\).

For a shelving unit with 6 type R shelves, there are 5 type S shelves.

For \(n\) type R shelves, there are \(n-1\) type S shelves.

The first term is \(1\) and the common difference is \(4\).

terms from the 41st to the 100th inclusive.

Divide by \(n\):

First term:

Let the first term be \(a\) and the common difference be \(d\).

The 6th term is 39: \[ a+5d=39 \]

The 19th term is 7.8: \[ a+18d=7.8 \]

Subtract: \[ 13d=-31.2 \]

Then \[ a+5(-2.4)=39 \]

The sum of the first 25 terms is \[ S_{25}=\frac{25}{2}\left(2a+24d\right) \]

So

The \(n\)th term is

terms from the 50th to the 100th inclusive.

Substitute the two term numbers into the given rule.

Divide both entries by their highest common factor, 5.

The common difference is \(8-3=5\), so the rule has the form \(5n+c\).

A constant difference gives a linear nth term.

At \(n=1\), \(5+c=3\), giving \(c=-2\). Thus \(u_n=5n-2\).

Use the first term to find the constant.

With first term \(3\) and difference \(8\), \[S_n=\frac n2[2(3)+(n-1)8]=\frac n2(8n-2)=4n^2-n.\]

Use the arithmetic-series sum formula with the given first term and common difference.

Substitute 5k and 3k as the term counts, then subtract the complete expressions.

\[32k^2-k-12780=0\quad\Rightarrow\quad(k-20)(32k+639)=0.\] The roots are \(k=20\) and \(k=-639/32\). Reject the negative root because the numbers of terms must be positive.

Factor the quadratic and apply the contextual restriction.

Use the nth-term formula with 4k=80, not with k=20.

**Step 1: Find the common difference *d*.** The *n*^th term is given by *u*_*n* = *a* + (*n*−1)*d*. *u*₂₀ = 10 + 19*d* = 86 19*d* = 76 $$d = \\frac{76}{19} = 4$$

**Step 2: Use the sum formula to form an equation in terms of *N*.** $$S_N = \\frac{N}{2} \\left\[2a + (N - 1)d \\right\]$$ Substitute the known values *S*_*N* = 5194, *a* = 10, and *d* = 4: $$5194 = \\frac{N}{2} \\left\[2(10) + (N - 1)(4) \\right\]$$

**Step 3: Simplify and solve the quadratic equation.** Multiply by 2 to clear the fraction: 10388 = *N*\[20+4*N*−4\] 10388 = *N*(4*N*+16) 4*N*² + 16*N* − 10388 = 0

Divide the entire equation by 4 to make it simpler: *N*² + 4*N* − 2597 = 0

**Step 4: Use the quadratic formula to find *N*.** $$N = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$ $$N = \\frac{-4 \\pm \\sqrt{16 - 4(1)(-2597)}}{2}$$ $$N = \\frac{-4 \\pm \\sqrt{16 + 10388}}{2}$$ $$N = \\frac{-4 \\pm \\sqrt{10404}}{2}$$ $$N = \\frac{-4 \\pm 102}{2}$$

Since *N* must be a positive integer, we discard the negative root: $$N = \\frac{-4 + 102}{2} = \\frac{98}{2} = 49$$

Double the first equation:

Subtract:

Let the first term be \(a\) and the common difference be \(d\).

The 8th term plus the 20th term is 284: \[ (a+7d)+(a+19d)=284 \]

Subtract the equations: \[ 3d=9 \]

Then \[ 2a+26(3)=284 \]

Now find \(S_{45}\): \[ S_{45}=\frac{45}{2}(2a+44d) \]

For an arithmetic sequence, consecutive differences are equal:

So the first term is

and the common difference is

The sum of the first \(n\) terms is

Only the positive value is possible:

So

The 8th term is 45:

Then

and

Subtract:

Since \(d\) is prime and \(m^2\) is a square factor,

So

Subtract:

Let the first term be \(a\) and the common difference be \(d\).

So

and

Given \(S_{10}=4S_5\):

The 8th term is \(45\):

Substitute \(d=2a\):

\(a=k+2\), \(d=3\).

The first term is \(a=k+2\) and the common difference is \(d=3\).

Successive listed terms differ by 3.

Apply the arithmetic-series sum formula.

Use 2n in place of n throughout the sum formula.

Using \(S_{2n}=4S_n\) and \(n>0\), \[2k+1+6n=2(2k+1+3n)=4k+2+6n.\]

Divide the equation by n, then expand the right-hand side.

The n-terms cancel, leaving a linear equation for k.

The 100th term is the first term plus 99 common differences.

For an arithmetic series,

Given \(S_9=216\):

Now solve

Then

Series P: first term *a*, c.d. *d*; *S*₄₀(*P*) = 2070. Series Q: first term 2*a*, c.d. 3*d*; 25th term of Q  = 186. Find *S*₄₀(*Q*).

Form equation from *S*₄₀(*P*) = 2070: $$S\_{40} = \\frac{40}{2}\[2a+39d\] = 20(2a+39d) = 2070 \\implies 2a + 39d = 103.5 \\tag{1}$$

Form equation from 25th term of Q  = 186: *T*₂₅(*Q*) = 2*a* + 24(3*d*) = 2*a* + 72*d* = 186

Solve simultaneously — subtract (1) from (2): 33*d* = 82.5 ⟹ *d* = 2.5

Substitute *d* = 2.5 into (1): 2*a* + 39(2.5) = 103.5 ⟹ 2*a* = 6 ⟹ *a* = 3

Compute *S*₄₀(*Q*), with first term  = 2*a* = 6 and c.d.  = 3*d* = 7.5: $$S\_{40}(Q) = \\frac{40}{2}\\bigl\[2(6) + 39(7.5)\\bigr\] = 20\\bigl\[12 + 292.5\\bigr\] = 20 \\times 304.5 = 6090$$

The interior angles form an arithmetic sequence with first term \(84\) and common difference \(4\).

So the sum of the angles is

The sum of the interior angles of an \(n\)-sided polygon is

So

Since \(n>5\),

Therefore the sum of the interior angles is

The common difference is \(4\), so the nth term is \[ 4n-1 \]

The last term is \(x\), so \[ x=4n-1 \]

The sum is 7260: \[ \frac{n}{2}(3+x)=7260 \]

Substitute \(n=\dfrac{x+1}{4}\): \[ \frac{x+1}{8}(x+3)=7260 \]

Factorise: \[ (x-239)(x+243)=0 \]

The last term is positive, so \[ x=239 \]

Double it:

Subtract:

Now

Let \(u_6=v\).

For an arithmetic series,

The average of \(u_1\) and \(u_{11}\) is \(u_6\), so

So

But \(u_4=6\). If \(u_6=2\), then the common difference is negative and later terms would not all be positive. Therefore,