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Mathematicians!!! Please Contribute To This Question by mostsimpliest(m): 1:24pm On Nov 23, 2016
The sum of the first ten terms of an AP is 15 and the sum of the next ten terms is 215,whats the sum of it 28th term?
Re: Mathematicians!!! Please Contribute To This Question by dejt4u(m): 1:36pm On Nov 23, 2016
mostsimpliest:
The sum of the first ten terms of an AP is 15 and the sum of the next ten terms is 215,whats the sum of it 28th term?
answer is 546.. a = -7.5 and d = 2
Re: Mathematicians!!! Please Contribute To This Question by dejt4u(m): 1:36pm On Nov 23, 2016
Re: Mathematicians!!! Please Contribute To This Question by NDSMELODY(m): 2:00pm On Nov 23, 2016
Pls show us the workings.......i dnt think dat question is complete.....becos forming an equation leads to simultenous equation which are dsame
Re: Mathematicians!!! Please Contribute To This Question by dejt4u(m): 2:16pm On Nov 23, 2016
NDSMELODY:
Pls show us the workings.......i dnt think dat question is complete.....becos forming an equation leads to simultenous equation which are dsame
the question is complete and correct.. The equations and solution will be posted
Re: Mathematicians!!! Please Contribute To This Question by dejt4u(m): 2:39pm On Nov 23, 2016
mostsimpliest:
The sum of the first ten terms of an AP is 15 and the sum of the next ten terms is 215,whats the sum of it 28th term?
sum of the first ten terms of an AP is 15;
S10 = 15;
Remember, sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S10 = 10/2 (2a + 9d)
but S10 = 15,
therefore, 10/2 (2a + 9d) = 15,
5(2a+9d) = 15,
2a + 9d = 15/5
2a + 9d = 3.............equation (i)

the sum of the next ten terms is 215:
this means that S20 - S10 = 215,
make S20 the subject of the formula,
S20 = 215 + S10
but r'mber that S10 = 15,
therefore,
S20 = 215 + 15,
S20 = 230,
again, sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S20 = 20/2 (2a + 19d) and Recall that S20 = 230,
so, 20/2 (2a + 19d) = 230,
10(2a + 19d) = 230,
2a + 19d = 23............equation (ii)

now combine eqn (i) and eqn (ii) together and solve for a and d simultaneously..

2a + 9d = 3.............eqn (i)
2a + 19d = 23............eqn (ii)

elimination method; eqn (ii) minus eqn (i),
0 + 10d = 20,
d = 20/10,
d = 2

subtitute for the value of d in eqn (i)
2a + 9d = 3.............eqn (i)
2a = 3 - 9d,
2a = 3 - 9(2),
2a = 3 - 18,
2a = -15,
a = -15/2,
a = -7.5

now a = -7.5 and d = 2..

The question; whats the sum of it 28th term?
S28 = ?
But sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S28 = 28/2 (2a + 27d),
S28 = 14 (2(-7.5) + 27(2)),
S28 = 14 (-15 + 54),
S28 = 14 (39)
S28 = 546 ......QED

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Re: Mathematicians!!! Please Contribute To This Question by NDSMELODY(m): 3:40pm On Nov 23, 2016
Where did u get 20
Re: Mathematicians!!! Please Contribute To This Question by dejt4u(m): 3:58pm On Nov 23, 2016
NDSMELODY:
Where did u get 20
the sum of the first 10 was given to you.. Then sum of the next 10 will be S20 - S10
Re: Mathematicians!!! Please Contribute To This Question by mostsimpliest(m): 5:29pm On Nov 23, 2016
dejt4u:

sum of the first ten terms of an AP is 15;
S10 = 15;
Remember, sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S10 = 10/2 (2a + 9d)
but S10 = 15,
therefore, 10/2 (2a + 9d) = 15,
5(2a+9d) = 15,
2a + 9d = 15/5
2a + 9d = 3.............equation (i)

the sum of the next ten terms is 215:
this means that S20 - S10 = 215,
make S20 the subject of the formula,
S20 = 215 + S10
but r'mber that S10 = 15,
therefore,
S20 = 215 + 15,
S20 = 230,
again, sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S20 = 20/2 (2a + 19d) and Recall that S20 = 230,
so, 20/2 (2a + 19d) = 230,
10(2a + 19d) = 230,
2a + 19d = 23............equation (ii)

now combine eqn (i) and eqn (ii) together and solve for a and d simultaneously..

2a + 9d = 3.............eqn (i)
2a + 19d = 23............eqn (ii)

elimination method; eqn (ii) minus eqn (i),
0 + 10d = 20,
d = 20/10,
d = 2

subtitute for the value of d in eqn (i)
2a + 9d = 3.............eqn (i)
2a = 3 - 9d,
2a = 3 - 9(2),
2a = 3 - 18,
2a = -15,
a = -15/2,
a = -7.5

now a = -7.5 and d = 2..

The question; whats the sum of it 28th term?
S28 = ?
But sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S28 = 28/2 (2a + 27d),
S28 = 14 (2(-7.5) + 27(2)),
S28 = 14 (-15 + 54),
S28 = 14 (39)
S28 = 546 ......QED

nycc1,oil dey ya head
Re: Mathematicians!!! Please Contribute To This Question by NL1960: 5:43pm On Nov 23, 2016
dejt4u:

sum of the first ten terms of an AP is 15;
S10 = 15;
Remember, sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S10 = 10/2 (2a + 9d)
but S10 = 15,
therefore, 10/2 (2a + 9d) = 15,
5(2a+9d) = 15,
2a + 9d = 15/5
2a + 9d = 3.............equation (i)

the sum of the next ten terms is 215:
this means that S20 - S10 = 215,
make S20 the subject of the formula,
S20 = 215 + S10
but r'mber that S10 = 15,
therefore,
S20 = 215 + 15,
S20 = 230,
again, sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S20 = 20/2 (2a + 19d) and Recall that S20 = 230,
so, 20/2 (2a + 19d) = 230,
10(2a + 19d) = 230,
2a + 19d = 23............equation (ii)

now combine eqn (i) and eqn (ii) together and solve for a and d simultaneously..

2a + 9d = 3.............eqn (i)
2a + 19d = 23............eqn (ii)

elimination method; eqn (ii) minus eqn (i),
0 + 10d = 20,
d = 20/10,
d = 2

subtitute for the value of d in eqn (i)
2a + 9d = 3.............eqn (i)
2a = 3 - 9d,
2a = 3 - 9(2),
2a = 3 - 18,
2a = -15,
a = -15/2,
a = -7.5

now a = -7.5 and d = 2..

The question; whats the sum of it 28th term?
S28 = ?
But sum of nth term
Sn = n/2 [2a + (n-1)d]
Then, S28 = 28/2 (2a + 27d),
S28 = 14 (2(-7.5) + 27(2)),
S28 = 14 (-15 + 54),
S28 = 14 (39)
S28 = 546 ......QED

Chai, this brings back fond memories of Additional Mathematics popularly called 'Addico' (now called Further Mathematics) of those days when only hot brains could hold the red coloured Additional Mathematics textbook. cool.

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