Sequences and Series

31 questions

1.25: Past-paper question 4

4PM1/1/November/2025 — Question 4 · 8 marks

Given that r=1n(4r+A)=n2(4n6)\displaystyle \sum_{r=1}^n (4r + A) = \frac{n}{2}(4n - 6)
where A\displaystyle A is a constant
(a) show that A=5\displaystyle A = -5
(3)
Hence, or otherwise,
(b) (i) evaluate 2160(4r5)\displaystyle \sum_{21}^{60} (4r - 5)
(2)
(ii) find the greatest value of n\displaystyle n such that
(3)
r=1n(4r5)<3418\displaystyle \sum_{r=1}^n (4r - 5) < 3418

1.23: Past-paper question 4

4PM1/2/June/2025 — Question 4 · 6 marks

A geometric series G\displaystyle G has first term a\displaystyle a and common ratio r\displaystyle r
The third term of G\displaystyle G is 10 and the seventh term of G\displaystyle G is 40
Given that the second term of G\displaystyle G is negative,
(a) find
(i) the exact value of r\displaystyle r
(ii) the value of a\displaystyle a
(5)
(b) Explain why G\displaystyle G does not have a sum to infinity.
(1)

1.26: Past-paper question 6

4PM1/2/November/2025 — Question 6 · 8 marks

(i) A geometric series G\displaystyle G has first term (3+3)\displaystyle (3 + \sqrt{3}) and common ratio r\displaystyle r
The second term of G\displaystyle G is 3\displaystyle \sqrt{3}
(a) Find, showing all your working, the exact value of r\displaystyle r
Give your answer in the form p+3q\displaystyle \frac{p + \sqrt{3}}{q} where p\displaystyle p and q\displaystyle q are integers to be found.
(2)
(b) Explain why G\displaystyle G is convergent.
(1)
(ii) A different geometric series H\displaystyle H has first term 8 and common ratio 0.6
The sum to n\displaystyle n terms of H\displaystyle H is Sn\displaystyle S_n and the sum to infinity of H\displaystyle H is T\displaystyle T
Find, using logarithms, the least value of n\displaystyle n such that TSn<0.12\displaystyle T - S_n < 0.12
(5)

1.24: Find the value of (ii) The common ratio of a geometric series is positive.

4PM1/2R/June/2025 — Question 4 · 11 marks

(i) Find the value of k=6173(2)k1\displaystyle \sum_{k=6}^{17} 3(2)^{k-1}
(4)
(ii) The common ratio of a geometric series G\displaystyle G is positive.
The third term of G\displaystyle G is 713\displaystyle \frac{7}{13} and the ninth term of G\displaystyle G is 4489477\displaystyle \frac{448}{9477}
G\displaystyle G is convergent with sum to infinity S\displaystyle S
Find the exact value of S\displaystyle S
(7)

1.16: Show that (b) Hence, or otherwise, evaluate Given that (c) find the value of

4PM1/1/June/2024 — Question 3 · 8 marks

(a) Show that r=1n(5r3)=n2(5n1)\displaystyle \sum_{r=1}^n (5r - 3) = \frac{n}{2}(5n - 1)
(3)
(b) Hence, or otherwise, evaluate r=3160(5r3)\displaystyle \sum_{r=31}^{60} (5r - 3)
(2)
Given that r=1n(5r3)=3783\displaystyle \sum_{r=1}^n (5r - 3) = 3783
(c) find the value of n\displaystyle n
(3)

1.17: Past-paper question 8

4PM1/1/June/2024 — Question 8 · 12 marks

The sum of the first and second terms of a geometric series G\displaystyle G is 400
The sum of the second and third terms of G\displaystyle G is 100
(a) Show that the common ratio of G\displaystyle G is 14\displaystyle \frac{1}{4}
(4)
(b) Show that the first term of G\displaystyle G is 320
(2)
(c) Find the sum to infinity of G\displaystyle G
(2)
The sum to n\displaystyle n terms of G\displaystyle G is Sn\displaystyle S_n
(d) Find, using logarithms, the least value of n\displaystyle n such that
Sn>426.6S_n > 426.6
(4)

1.20: Past-paper question 2

4PM1/1/November/2024 — Question 2 · 13 marks

The sum of the fifth, sixth and seventh terms of an arithmetic series A\displaystyle A is nine times the sum of the first and second terms.
The third term of A\displaystyle A is 12
(a) Find the first term and common difference of A\displaystyle A
(5)
The n\displaystyle nth term of A\displaystyle A is un\displaystyle u_n
(b) Find the value of r=1560ur\displaystyle \sum_{r=15}^{60} u_r
(4)
The sum to n\displaystyle n terms of A\displaystyle A is Sn\displaystyle S_n
Given that 2Sn5un=10\displaystyle 2S_n - 5u_n = 10
(c) find the value of n\displaystyle n
(4)

1.18: The sum of the first 10 terms of an arithmetic series is where is a constant.

4PM1/1R/June/2024 — Question 5 · 11 marks

The sum of the first 10 terms of an arithmetic series A\displaystyle A is 36k+1\displaystyle 36k + 1 where k\displaystyle k is a constant.
The 6th term of A\displaystyle A is 4k+1\displaystyle 4k + 1
(a) (i) Find an expression in terms of k\displaystyle k for the common difference of A\displaystyle A
(ii) Show that the first term of A\displaystyle A is 8\displaystyle -8
(5)
Given that the 4th term of A\displaystyle A is 7
(b) show that k=4\displaystyle k = 4
(2)
The sum of the first n\displaystyle n terms of A\displaystyle A is Sn\displaystyle S_n and the n\displaystyle nth term of A\displaystyle A is Un\displaystyle U_n
(c) Find the value of n\displaystyle n such that Sn=5Un+10+105\displaystyle S_n = 5U_{n+10} + 105
(4)

1.21: Past-paper question 7

4PM1/2/November/2024 — Question 7 · 16 marks

(a) Use the factor theorem to show that (4x1)\displaystyle (4x - 1) is a factor of
f(x)=64x364x2+3f(x) = 64x^3 - 64x^2 + 3
(2)
(b) Hence, or otherwise, find the exact roots of the equation
f(x)=0f(x) = 0
(4)
A geometric series G\displaystyle G has first term a\displaystyle a and common ratio r\displaystyle r
The third term of G\displaystyle G is 9 and the sum to infinity of G\displaystyle G is 192
(c) Show that 64r364r2+3=0\displaystyle 64r^3 - 64r^2 + 3 = 0
(3)
Given that r\displaystyle r is a rational number
(d) write down the value of r\displaystyle r
(1)
(e) show that a=144\displaystyle a = 144
(2)
The sum to n\displaystyle n terms of G\displaystyle G is Sn\displaystyle S_n
(f) Using logarithms, find the least value of n\displaystyle n such that Sn>191.9\displaystyle S_n > 191.9
(4)

1.19: Past-paper question 8

4PM1/2R/June/2024 — Question 8 · 14 marks

The sum of the first 2 terms of a geometric series G\displaystyle G is 360
The sum of the 2nd and 3rd terms of G\displaystyle G is 288
The n\displaystyle nth term of G\displaystyle G is Un\displaystyle U_n
(a) Show that Un=A(45)n1\displaystyle U_n = A \left( \frac{4}{5} \right)^{n-1} where A\displaystyle A is an integer to be found.
(7)
(b) Explain why G\displaystyle G is convergent.
(1)
(c) Hence find the sum to infinity of G\displaystyle G
(2)
(d) Find the least number of terms for which the sum is greater than 978
(4)

1.11: Past-paper question 1

4PM1/1/June/2023 — Question 1 · 5 marks

(a) Show that r=1n(3r+2)=n2(3n+7)\quad (a) \text{ Show that } \sum_{r=1}^n (3r + 2) = \frac{n}{2}(3n + 7)
(3)
(b) Hence, or otherwise, evaluate r=1040(3r+2)(b) \text{ Hence, or otherwise, evaluate } \sum_{r=10}^{40} (3r + 2)
(2)

1.14: Past-paper question 7

4PM1/1/November/2023 — Question 7 · 9 marks

A geometric series G\displaystyle G with common ratio r\displaystyle r, has first term 16 and third term 2704625\displaystyle \frac{2704}{625}
(a) Find the two possible values of r\displaystyle r
(2)
Given that r>0\displaystyle r > 0
(b) find the sum to infinity of G\displaystyle G
(2)
The sum to n\displaystyle n terms of G\displaystyle G is greater than 33
(c) Find, using logarithms, the least possible value of n\displaystyle n
Show your working clearly.
(5)

1.12: Past-paper question 8

4PM1/1R/June/2023 — Question 8 · 10 marks

The n\displaystyle nth term of a geometric series G\displaystyle G is Un\displaystyle U_n and the sum of the first n\displaystyle n terms of G\displaystyle G is Sn\displaystyle S_n
Given that Un=254(35)n\displaystyle U_n = \frac{25}{4} \left( \frac{3}{5} \right)^n
(a) find the exact value of U5\displaystyle U_5
(1)
(b) Show that Sn=r=1nAB(35)r1\displaystyle S_n = \sum_{r=1}^n \frac{A}{B} \left( \frac{3}{5} \right)^{r-1} where A\displaystyle A and B\displaystyle B are integers to be found.
(3)
The sum to infinity of G\displaystyle G is S\displaystyle S
(c) Find the least value of n\displaystyle n such that SSn<0.045\displaystyle S - S_n < 0.045
(6)

1.13: The th term of a convergent geometric series is Find the sum to infinity of the series.

4PM1/2/June/2023 — Question 2 · 6 marks

The n\displaystyle nth term of a convergent geometric series is 8(12n)\displaystyle 8^{(1-2n)}
Find the sum to infinity of the series.
Give your answer in the form pq\displaystyle \frac{p}{q} where p\displaystyle p and q\displaystyle q are integers to be found.
(6)

1.15: Past-paper question 8

4PM1/2/November/2023 — Question 8 · 10 marks

The sum to n\displaystyle n terms of an arithmetic series A\displaystyle A is Sn\displaystyle S_n
The sum of the first four terms of A\displaystyle A is 42 and the fifth term of A\displaystyle A is 23
(a) Show that Sn=r=1n(PrQ)\displaystyle S_n = \sum_{r=1}^n (Pr - Q) where P\displaystyle P and Q\displaystyle Q are prime numbers.
(6)
S2n3Un=1062\displaystyle S_{2n} - 3U_n = 1062 where Un\displaystyle U_n is the n\displaystyle nth term of A\displaystyle A
(b) Find the value of n\displaystyle n
(4)

1.1: Finite and infinite sums of a geometric series

4PM1/1/June/2022 — Question 4 · 7 marks

The common ratio of a geometric series G\displaystyle G is positive.
The sum of the first 4 terms of G\displaystyle G is 80\displaystyle 80.
The sum to infinity of G\displaystyle G is 81\displaystyle 81.
Show that the sum of the first 7 terms of G\displaystyle G differs from the sum to infinity of G\displaystyle G by
127.\frac{1}{27}.
(7)

1.3: A geometric series with a surd sum

4PM1/2R/June/2022 — Question 3 · 6 marks

A geometric series has first term a\displaystyle a and common ratio r\displaystyle r, where r>0\displaystyle r>0.
Given that the 3rd term of the series is 5\displaystyle 5 and that the 5th term of the series is 52\displaystyle \frac{5}{2},
(a) find
(i) the exact value of r\displaystyle r,
(ii) the value of a\displaystyle a.
(4)
(b) Find the sum to infinity of this series. Give your answer in the form
p+q2,p+q\sqrt{2},
where p\displaystyle p and q\displaystyle q are integers.
(2)

1.4: Sum of an arithmetic series in logarithmic form

4PM1/1/June/2021 — Question 4 · 5 marks

The n\displaystyle nth term of an arithmetic series is un\displaystyle u_n, where
un=(n+1)ln4.u_n=(n+1)\mathrm{ln}\,4.
Given that the sum of the first n\displaystyle n terms of the series is Sn\displaystyle S_n, show that
Sn=ln2(n2+an),S_n=\mathrm{ln}\,2^{\left(n^2+an\right)},
where a\displaystyle a is an integer whose value is to be found.
(5)

1.5: A convergent geometric series

4PM1/2/June/2021 — Question 5 · 7 marks

The n\displaystyle nth term of a geometric series with common ratio r\displaystyle r is un\displaystyle u_n.
Given that
u2+u4=212.5andu3+u4=62.5,u_2+u_4=212.5 \qquad \text{and} \qquad u_3+u_4=62.5,
(a) find the two possible values of r\displaystyle r.
(5)
Given that the series is convergent with sum to infinity S\displaystyle S,
(b) find the exact value of S\displaystyle S.
(2)

1.6: An arithmetic series and its sums

4PM1/1/November/2020 — Question 6 · 10 marks

An arithmetic series A\displaystyle A has first term a\displaystyle a and common difference d\displaystyle d. The sum Sn\displaystyle S_n of the first n\displaystyle n terms of A\displaystyle A is given by
Sn=n(15+2n).S_n=n(15+2n).
(a) Find the value of a\displaystyle a and the value of d\displaystyle d.
(4)
(b) Find the 20th term of A\displaystyle A.
(2)
Given that
S2p2Sp=1+Sp1,S_{2p}-2S_p=1+S_{p-1},
(c) find the value of p\displaystyle p.
(4)

1.7: An arithmetic series of logarithms

4PM1/1R/November/2020 — Question 3 · 7 marks

The n\displaystyle nth term of an arithmetic series is un\displaystyle u_n such that
un=lna+(n1)lnb,u_n=\mathrm{ln}\,a+(n-1)\mathrm{ln}\,b,
where a\displaystyle a and b\displaystyle b are positive integers.
Given that u2=ln12\displaystyle u_2=\mathrm{ln}\,12 and that u5=ln768\displaystyle u_5=\mathrm{ln}\,768, find the value of a\displaystyle a and the value of b\displaystyle b.
(7)

1.8: Evaluate sums using sigma notation

4PM1/2/November/2020 — Question 5 · 8 marks

(a) Show that
r=1n(3r+5)=12n(3n+13).\sum_{r=1}^{n}(3r+5)=\frac12n(3n+13).
(3)
(b) Hence evaluate
r=3550(3r+5).\sum_{r=35}^{50}(3r+5).
(2)
Given that
r=1n(3r+5)=385,\sum_{r=1}^{n}(3r+5)=385,
(c) find the value of n\displaystyle n.
(3)

1.9: A geometric series and a partial-sum ratio

4PM1/2/November/2020 — Question 7 · 12 marks

A geometric series has first term (x3)\displaystyle (x-3), second term (x+1)\displaystyle (x+1) and third term (4x2)\displaystyle (4x-2).
(a) Find the two possible values of x\displaystyle x.
(5)
Given that x<1\displaystyle x<1,
(b) show that the series is convergent.
(2)
The sum to infinity of the series is S\displaystyle S.
(c) Find the value of S\displaystyle S.
(2)
The sum of the first n\displaystyle n terms of the series is Sn\displaystyle S_n. Given that
SSn=256255,\frac{S}{S_n}=\frac{256}{255},
(d) find the value of n\displaystyle n.
(3)

1.10: Related arithmetic and geometric series

4PM1/2R/November/2020 — Question 1 · 6 marks

The n\displaystyle nth term of an arithmetic series A\displaystyle A is an\displaystyle a_n. The n\displaystyle nth term of a geometric series G\displaystyle G is tn\displaystyle t_n.
For these two series,
a1=t1,a10=t3=48,a10=4t2.a_1=t_1, \qquad a_{10}=t_3=48, \qquad a_{10}=4t_2.
Find
(i) the common ratio of G\displaystyle G,
(ii) the common difference of A\displaystyle A.
(6)

1.30: Arithmetic Series and Term Relations

4PM1/1/January/2019 — Question 2 · 8 marks

The sum of the first n\displaystyle n terms of an arithmetic series is Sn\displaystyle S_n.
Given that
Sn=r=1n(4r+1),S_n=\sum_{r=1}^{n}(4r+1),
(a) show that
Sn=n(3+2n).S_n=n(3+2n).
(4)
The r\displaystyle rth term of this arithmetic series is tr\displaystyle t_r.
Given that
Sn+3=Sn+3t15,S_{n+3}=S_n+3t_{15},
(b) find the value of n\displaystyle n.
(4)

1.27: Arithmetic Series and an Inequality

4PM1/1R/June/2019 — Question 6 · 10 marks

(a) Show that
r=1n(4r3)=n(2n1).\sum_{r=1}^{n}(4r-3)=n(2n-1).
(3)
(b) Hence, or otherwise, find the least value of n\displaystyle n such that
r=1n(4r3)>1000.\sum_{r=1}^{n}(4r-3)>1000.
(3)
Given that Sn=n(2n1)\displaystyle S_n=n(2n-1), tn=4n3\displaystyle t_n=4n-3 and that
18+3tn+7=Sn+4,18+3t_{n+7}=S_{n+4},
(c) find the value of n\displaystyle n.
(4)

1.28: The sum of the first terms of an arithmetic series is An where

4PM1/2/June/2019 — Question 7 · 9 marks

The sum of the first n\displaystyle n terms of an arithmetic series is An where
A ln=r=1n(4r+5)\displaystyle {\mathit{l}}_{n}\,=\,\sum_{r=1}^{n}\,(4r^{\star}+5)
(a) For this arithmetic series, find
(i) the first term,
(ii) the common difference.
(2)
The sum of the first n\displaystyle n terms of a geometric series is Gn where
Gn =r=1n4(3)r1\displaystyle =\sum_{r=1}^{n}4(3)^{r-1}
(b) For this geometric series, find
(i) the first term,
(ii) the common ratio.
(2)
(c) Find the value of n\displaystyle n for which A14 6=\displaystyle - 6 = Gn
(5)

1.31: The A geometric nth term series of the has series first is term Un a and common ratio

4PM1/2/January/2019 — Question 10 · 10 marks

The A geometric nth term series of the has series first is term Un a and common ratio r(r>0)\displaystyle r (r > 0)
Given that U1+3U2=8\displaystyle U1 + 3U2 = 8 and that U2×U3=4U5\displaystyle U2 \times U3 = 4U5
(a) find
(i) the value of r\displaystyle r
(ii) the value of a
(5)
(b) Hence show that Un =2n+23n\displaystyle ={\frac{2^{n+2}}{3^{n}}}
(2)
(c) Find the least value of n\displaystyle n such that Un <0.05\displaystyle < 0.05
(3)

1.29: The nth term of a geometric series is un

4PM1/2R/June/2019 — Question 7 · 11 marks

The nth term of a geometric series G\displaystyle G is un
The first term of G\displaystyle G is a and the common ratio of G\displaystyle G is J\displaystyle {\mathit{J}}_{\cdot} where r0\displaystyle r\geqslant0
Given that u3=4\displaystyle u3 = 4 and that u7=16\displaystyle u7 = 16
(a) (i) show that U=2\displaystyle U = 2
(ii) find the value of a.
(3)
(b) Find the least value of n\displaystyle n for which un 500\displaystyle 500
(4)
The sum of the first n\displaystyle n terms of G\displaystyle G is Sn
(c) Find Give your answer in the form p(1+2)\displaystyle p (1 + 2) where p\displaystyle p is an integer.
(4)