Mathematics-- Solution to exercises on previous tutorial
Introduction
It's me again, Hayzeed, the mathematics and electrical courses blogger, today we are solving the exercises on the last tutorial which is on Number sequence
- Determine (a) the 9th, and (b) the
16th term of the series 2, 7, 12, 17,...
From the question, we are to determine the 9th and 16th term.
The series given is:** 2, 7, 12, 17**
Having looked at the series, the common difference d is 5 and the first term is 2.
How did we got that?
It's just by subtracting 1st term from the 2nd term or 2nd term from the 3rd term.
Okay, the formula for finding nth term of an AP is given by : Tn= a + (n - 1)d
where a= first term=2
n= number of term
d= common difference=5
(a) For the 9th term
n= 9
Then T9= 2 + (9 - 1)5
=2 + (8)5= 2 +40= 42
So the 9th term of the series is 42.
Is it clear?
If yes, drop your phone, take out your pen and book then find the 16th term of the series.
(b) For the 16th term
n=16
Then T16= 2 +( 16 -1)5
= 2 +(17)5 = 2 + 85= 87.
So the 16th term of the series is 87.
Do you also get that as well?
(2) The 6th term of an AP is 17 and the
13th term is 38. Determine the 19th term.
You can see that we are not given series this time and the first term and the common difference is not stated in the question. Before we can know the 19th term, we need to determine the first term and the common difference..
From the first statement, " the 6th term of an AP is 17", this implies that the 6th term is equal to 17.
From the formula Tn= a + (n - 1)d
n= 6. T6=17
Substituting that into the formula.
We have, a +(6 -1)d=17
a + 5d=17 ----------------------- (1)
By repeating the same process for the 2nd statement.
We have, a + ( 13 - 1)d= 38
a + 12d =38 -----------------------(2)
Solving (1) and (2) simultaneously using elimination method.
subtracting (1) and (2)
a + 5d -(a + 12d) = 17 -38
a + 5d - a - 12d = -21
5d - 12d = -21
- 7d = -21
Dividing both side by -7
We have , d= -21/-7 = 3
Substituting the value of d in (1)
a + 5(3)= 17
a + 15=17
a= 17 -15=2
Hence, first term a= 2
Common difference d=3.
Then the 19th term can be determined using the nth term formula which is Tn= a +(n -1)d
So, the 19th term= 2 + (19-1)3
=2 +(18)3= 2 + 54= 56.
The 19th term of the AP is 56.
Do you understand??
If yes, can you solve the third question in the exercise?
(3) Determine the number of the term which is 29
in the series 7, 9.2, 11.4, 13.6, ...
We are to determine the number of the term that will be equal to 29. I.e a +(n - 1)d =29
First term a=7
Common difference d= 9.2 -7=11.4 - 9.2= 2.2
Then, we have , 7 + (n -1)2.2=29
7 + 2.2n - 2.2=29
2.2n + 4.8= 29
2.2n= 29 - 4.8
2.2n= 24.2
n= 24.2/2.2=11
So,the number of terms is 11....
We are continuing the tutorial on the next tutorial_
References
Engineering mathematics by John Bird
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