Solutions to jamb mathematics sequences and series questions: Jamb mathematics tutorial

Hi guys. Today on jamb mathematics tutorials, we’ll be taking a look at some selected jamb past questions on sequences and series and providing solutions to such questions. Feel free to drop any question giving you tough time on the comment section.

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1993-23: If k+1,2k-1,3k+1 are three consecutive terms of a geometric progression, find the possible values of the common ratio.

Solution

r=\frac{2k-1}{k+1}=\frac{3k+1}{2k-1}\\\frac{2k-1}{k+1}=\frac{3k+1}{2k-1}\\(2k-1)(2k-1)=(3k+1)(k+1)\\4k^2-2k-2k+1=3k^2+3k+k+1\\4k^2-4k+1=3k^2+4k+1\\k^2-8k=0\\k(k-8)=0\\k=0 or k-8=0\\k=0 or k=8

substitute k=0 or k=8 into r=\frac{2k-1}{k+1}

K=0, r=\frac{-1}{1}=-1

if k=8, r=\frac{16-1}{8+1}=\frac{15}{9}=\frac{5}{3}\\\therefore r=-1, \frac{5}{3}

 

1994-19: If the 6th term of an arithmetic progression is 11 and the first term is 1, find the common difference

Solution

a=1\cdots eq(i)

a+5d=11\cdots eq(ii)

substitute a=1 into eq(ii)

1+5d=11\\5d=10\\d=2

 

1995-17: The 4_{th} term of an A.P is 13 while the 10_{th} term is 31. Find the 21_{st} term.

Solution

a+3d=13\cdots eq(i)

a+9d=31\cdots eq(ii)

Subtract eq(i) from eq(ii)

6d=18\\d=3

substitute d=3 into eq(i)

a+3(3)=13\\a+9=13\\a=4\\U_{21}=a+20d\\U_{21}=4+20(3)\\U_{21}=64

 

1995-23: Find the sum to infinity of the following sequence 1,\frac{9}{10},(\frac{9}{10})^2,(\frac{9}{10})^3.

Solution

a=1\cdotseq(i)

ar=\frac{9}{10}\cdotseq(ii)

divide eq(ii) by eq(i)

r=\frac{9}{10}\\S_{\infty}=\frac{a}{1-r}\\S_{\infty}=\frac{1}{1-\frac{9}{10}}\\S_{\infty}=\frac{1}{\frac{1}{10}}\\S_{\infty}=10

 

1997-20: The n_{th} term of a sequence is given by 3^{1-n}. Find the sum of the first three terms of the sequence.

Solution

U_{n}=3^{1-n}\\U_{1}=3^{1-1}=3^0=1\\U_{2}=3^{1-2}=3^{-1}=\frac{1}{3}\\U_{3}=3^{1-3}=3^{-2}=\frac{1}{9}\\S_{3}=U_{1}+U_{2}+U_{3}\\S_{3}=1+\frac{1}{3}+\frac{1}{9}\\S_{3}=\frac{13}{9}=1\frac{4}{9}

 

1998-19: If p+1,2p-10,1-4p^2 are three consecutive terms of an arithmetic progression, find the possible values of p.

Solution

d=(2p-10)-(p+1)=(1-4p^2)-(2p-10)\\(2p-10)-(p+1)=(1-4p^2)-(2p-10)\\2p-10-p-1=1-4p^2-2p+10\\p-11=11-4p^2-2p\\4p^2+3p-22=0\\(4p^2+11p)-(8p+22)=0\\p(4p+11)-2(4p+11)=0\\(p-2)(4p+11)=0\\p-2=0 or 4p+11=0\\p=2 or p=\frac{-11}{4}

 

1998-20: The sum of the first three terms of a geometric progression is half its sum to infinity. Find the possible common ratio of the progression.

Solution

S_{3}=\frac{a(1-r^3)}{1-r}\\S_{\infty}=\frac{a}{1-r}\\S_{3}=\frac{S_{\infty}}{2}\\\frac{a(1-r^3)}{1-r}=\frac{1}{2}(\frac{a}{1-r})\\2(1-r^3)=1\\(1-r)^3=\frac{1}{2}\\r^3=\frac{1}{2}\\r=\frac{1}{\sqrt[3]{2}}

 

1978-27}: In a geometric progression, the first term is 153 and the sixth term is \frac{17}{27}. The sum of the first four terms is?

Solution

a=153\cdots eq(i)

ar^5=\frac{17}{27}\cdots eq(i)

dividing eq(ii) by eq(i)

\frac{ar^5}{a}=\frac{\frac{17}{27}}{153}\\r^5=\frac{1}{243}\\r=(\frac{1}{243})^{\frac{1}{5}}\\r=\frac{1}{3}\\S_{n}=\frac{a(1-r^n)}{1-r}\\S_{4}=\frac{153(1-(\frac{1}{3})^4)}{1-\frac{1}{3}}\\S_{4}=\frac{153(1-\frac{1}{81})}{\frac{2}{3}}\\S_{4}=\frac{153\times 3\times \frac{80}{81}}{2}\\S_{4}=\frac{680}{3}

 

1978-28: An arithmetic progression has first term as 11 and fourth term as 32. The sum of the first nine terms is?

Solution

a=11\cdots eq(i)

a+3d=32\cdots eq(ii)

substitute a=11 into eq(ii)

11+3d=32\\3d=21\\d=7\\S_{n}=\frac{n}{2}(2a+(n-1)d)\\S_{9}=\frac{9}{2}(2(11)+(9-1)7)\\S_{9}=4\cdot5(22+8\times7)\\S_{9}=4\cdot5(22+56)\\S_{9}=4\cdot5(78)\\S_{9}=351

 

1981-38: The first term of an arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81.

Solution

a=3\cdots eq(i)

a+4d=9\cdots eq(ii)

substitute a=3 into eq(ii)

3+4d=9\\4d=6\\d=\frac{3}{2}\\S_{n}=\frac{n}{2}(2a+(n-1)d)\\S_{n}=81\\81=\frac{n}{2}(2(3)+(n-1)\frac{3}{2})\\162=n(6+1\cdot5(n-1))\\162=6n+1\cdot5n^2-1\cdot5n

divide through by 1\cdot5\\108=4n+n^2-n\\n^2+3n-108=0\\n^2 +12n-9n-108=0\\(n^2+12n)-(9n+108)=0\\n(n+12)-9(n+12)=0\\(n+12)(n-9)=0\\n+12=0, or n-9=0\\n=-12, or n=9

since the number of terms cannot be negative

n=9

 

{1987-29}: Find the sum of the first 21 terms of the progression -10,-8,-6,\ldots

Solution

from the progression, a=-10, d=-8-(-10)=2\\S_{n}=\frac{n}{2}(2a+(n-1)d)\\S_{21}=\frac{21}{2}(2(-10)+(21-1)2)\\S_{21}=10\cdot5(-20+(20)2)\\S_{21}=10\cdot5(-20+40)\\S_{21}=10\cdot5(20)\\S_{21}=210

 

1987-30}: Find the eleventh term of the progression 4,8,16,\cdots

Solution

a=4\\r=\frac{8}{4}=2

eleventh term =ar^{10}\\4\times2^{10}\\2^2\times2^{10}\\2^{12}

 

1988-28: If the sum of the 8_{th} and 9_{th} term of an arithmetic progression is 72 and the fourth term is -6, find the common difference.

Solution

U_{8}=a+7d, U_{9}=a+8d\\a+7d+a+8d=72\\2a+15d=72\cdots eq(i)

U_{4}=a+3d\\a+3d=-6\cdots eq(ii)

from eq(ii), a=-6-3d

substitute a=-6-3d into eq(i)

2(-6-3d)+15d=72\\-12-6d+15d=72\\9d=84\\d=9\frac{1}{3}

 

1991-29: Find the sum of the 20 terms in an arithmetic progression whose first term is 7 and last term is 117

Solution

a=7\\l=117 (where l is the last term)

n=20\\S_{n}=\frac{n}{2}(a+l)\\S_{20}=\frac{20}{2}(7+117)\\S_{20}=10(124)\\S_{20}=1240

 

1992-18: Find the sum to infinity of the following series 3+2+\frac{4}{3}+\frac{8}{9}+\frac{16}{27}+\ldots

Solution

S_{\infty}=\frac{a}{1-r}\\a=3\cdotseq(i)

ar=2\cdotseq(ii)

substitute a=3 into eq(ii)

3r=2\\r=\frac{2}{3}\\S_{\infty}=\frac{3}{1-\frac{2}{3}}\\S_{\infty}=\frac{3}{\frac{1}{3}}\\S_{\infty}=9

 

1992-20: For an arithmetic sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms.

Solution

a=2\\d=3\\S_{n}=\frac{n}{2}(2a+(n-1)d)\\S_{11}=\frac{11}{2}(2(2)+(11-1)3)\\S_{11}=\frac{11}{2}(4+(10)3)\\S_{11}=\frac{11}{2}(4+30)\\S_{11}=\frac{11}{2}(34)\\S_{11}=187

 

1988-29: If 7 and 189 are the first and fourth terms of a geometric progression respectively, find the sum of the first three terms of the progression.

Solution

a=7\cdots eq(i)

ar^3=189\cdots eq(ii)

divide eq(ii) by eq(i)

r^3=27\\r=3\\S_{n}=\frac{a(r^n-1)}{r-1}\\S_{3}=\frac{7(3^3-1)}{3-1}\\S_{3}=\frac{7(26)}{2}\\S_{3}=91

 

{1989-30}: If -8,m,n,19 are in arithmetic progression, find (m,n)

Solution

a=-8\\d=m-(-8)=n-m=19-n\\d=m+8=n-m=19-n\\m+8=n-m\\2m=n-8\\2m-n=-8\cdots eq(i)

m+8=19-n\\m+n=11\cdots eq(ii)

adding eq(i) to eq(ii)

3m=3\\m=1

substitute m=1 into eq(ii)

1+n=11\\n=10\\\therefore m=1,n=10

 

1990-27: Find the sum of the first twenty terms of the arithmetic progression \log{a},\log{a^2},\log{a^3},\ldots

Solution

a=\log a\cdots eq(i)

a+d=\log{a^2}\cdots eq(ii)

substitute a=\log a into eq(ii)

\log{a}+d=\log{a^2}\\d=\log{a^2}-\log{a}\\d=\log{\frac{a^2}{2}}\\d=log{a}\\S_{n}=\frac{n}{2}(2a+(n-1)d)\\S_{20}=\frac{20}{2}(2\log a+(20-1)\log a)\\S_{20}=10(2\log a+19\log a)\\S_{20}=10(21\log a)\\S_{20}=210\log a\\S_{20}=\log {a^{210}}

 

1999-17}: The first term of a geometric progression is twice its common ratio. Find the sum of the first two terms of the progression if its sum to infinity is 8

Solution

a=2r\\S_{\infty}=\frac{a}{1-r}

substitute a=21 into S_{\infty}\\S_{\infty}=\frac{2r}{1-r}\\8=\frac{2r}{1-r}\\8(1-r)=2r\\8-8r=2r\\10r=8\\r=\frac{8}{10}=\frac{4}{5}\\\therefore a=2(\frac{4}{5})=\frac{8}{5}\\S_{2}=\frac{a(1-r^2)}{1-r}\\S_{2}=\frac{\frac{8}{5}(1-(\frac{4}{5})^2)}{1-\frac{4}{5}}\\S_{2}=\frac{\frac{8}{5}(1-(\frac{16}{25}))}{\frac{1}{5}}\\S_{2}=\frac{\frac{8}{5}(\frac{9}{25})}{\frac{1}{5}}\\S_{2}=\frac{72}{25}=2\frac{22}{25}

 

2000-20}: The third term of an A.P is 4x-2y and the 9_{th} term is 1ox-8y. Find the common difference.

Solution

a+2d=4x-2y\cdots eq(i)

a+8d=10x-8y\cdots eq(ii)

subtract eq(i) from eq(ii)

6d=(10x-8y)-(4x-2y)\\6d=10x-8y-4x+2y\\d=x-y

 

2001-10: The sixth term of an arithmetic progression is half of its twelfth term. The first term is equal to ?

Solution

a+5d=\frac{1}{2}(a+11d)\\2(a+5d)=a+11d\\2a+10d=a+11d\\a=d

 

2002-29: If the 9th term of an A.P is five times the 5th term, find the relationship between a and d.

Solution

a+8d=5(a+4d)\\a+8d=5a+20d\\4a+12d=0\\a+3d=0

 

2003-16: Three consecutive terms of a geometric progression are given as n-2,n and n+3. Find the common ratio.

Solution

r=\frac{n}{n-2}=\frac{n+3}{n}\\\frac{n}{n-2}=\frac{n+3}{n}\\n^2=(n-2)(n+3)\\n^2=n^2+3n-2n-6\\n^2=n^2+n-6\\n-6=0\\n=6\\r=\frac{n}{n-2}\\r=\frac{6}{6-2}\\r=\frac{6}{4}=\frac{3}{2}=1\frac{1}{2}

 

2003-21: The sum of the first n terms of an arithmetic progression is 252. If the first term is -16 and the last term is 72, find the number of terms in the series

Solution

S_{n}=252\\a=-16\\l=72\\S_{n}=\frac{n}{2}(a+l)\\252=\frac{n}{2}(-16+72)\\504=n(56)\\n=9

 

2004-18: The n_{th} term of two sequences are Q_{n}=3\cdot2^{n-2} and U_{m}=3\cdot2^{2m-3}. Find the product of Q_{2} and U_{2}.

Solution

Q_{n}=3\cdot2^{n-2}\\Q_{2}=3\cdot2^{2-2}\\Q_{2}=3\cdot2^{0}\\Q_{2}=3\\U_{m}=3\cdot2^{2m-3}\\U_{2}=3\cdot2^{2(2)-3}\\U_{2}=3\cdot2^{4-3}\\U_{2}=3\cdot2^{1}\\U_{2}=6\\Q_{2}U_{2}=3\times6\\Q_{2}U_{2}=18

 

That’s the much we can take on this. If you have any question on this topic or related to mathematics, feel free to use the comment session. Be rest assured, there are mathematicians here. Also, notify us if you see any errors in the solutions above. Kindly share.

 

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