Arithmetic Progression is one of the topics in mathematics that sometime looks difficult to students. However, the challenge faced by the students may result from the fact that they are not properly taught.
The good news is that I am here to unveil the secret behind problems involving Arithmetic Progression. First, do me a favour and get your writing materials and probably your calculator and let’s get started.
The first term of any A.P is represented by “a” while the common difference is represented by “d”. if a set of numbers 2,4,6,8 …… are given, the first term is 2.
The common difference is the new term minus the previous term i.e 4-2, = 2, 6-4 = 2, 8-6 = 2 …
Therefore a Sequence of an A.P can be written thus a, a+d, a+rd, a+3d a+4d…
From the above, it is noted that the number of common difference must be added to a particular term in order to get the next term is less than the term being sought.
Given that an nth term is given by Tn
Then Tn = a+ (n-1) d
This is the formula for finding an nth term of an arithmetic progression.
See:
Write down the formula in your book for reference purposes. Having known the general formula for solving arithmetic progression, let’s get started by solving problems. Hope you are ready!
Solution to nth Term of Arithmetic Progression
Example 1: find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7
Solution
Get your data ready i.e
The first term a = 6
Common difference d = 7
Recall: the formula for nth term of an arithmetic progression
Tn = a+(n-1)d
The 6th term means that n=6
Let substitute the terms we have
T6 = 6+(6-1)x7
So we get rid of the bracket, remember BODMAS?
T6 = 6+5×7
Up next! Multiplication
i.e 5 multiply by 7
T6 = 6+35 = 41
Therefore T6 = 41
Similarly,
The 15th terms means n = 15
So,
T15 = 6+14×7
T15 = 6+98 = 104
Therefore T15 = 104
Therefore, T6 and T15 are 41 and 104 respectively
Example 2: if 8, A, B, C, D 38, 44, 50 …. Provides and A.P. find the values of A,B,C and D
Solution
The terms given are 8, A, B, C, D, 38, 44, 50…
The first term is 8
Similarly, the second term is A
The third term is B
The fourth term is C
The fifth term is D
The sixth term is 38
Recall: nth term of A.P
Tn = a+(n-1)d
i.e T6 = a+(6-1)d
a + 5d = 38 —- (1)
Substitute a=8 into equation 1
8+5xd = 58
8+5d = 38
Collect like terms that means the whole numbers to the right hand side
5d = 38-8
You wonder why the sign changed?
N/B: when a member crosses the equal to sign, the sign changes
5d = 30
Divide both sides by
5 (the coefficient of d)
5d/5 = 30/5
d= 6
Second term means n = 2
Using the nth term formula we then substitute what we have
Tn = a+(n-1)d
T2 = a+(2-1)d = A
a+(1)xd = A
but a=8 and d=6
8+1×6=A
A=14
Solving for B, takes the first method
T3 = a+(3-1)d = B
a+2d=B
8+2×6=B
8+12=B
B = 20
Again, T4 = a+(4-1)d = C
a+3d=C
8+3×6=C
8+18=C
C=26
Lastly, T5 = a+(5-1)d = D
a+4d=D
8+4×6=D
8+24=D
D=32
Therefore values of A,B,C, and D are 14,20,26 and 32 respectively
Example
The first term of an A.P is 11 while the last term is 144. If the common difference is 7. Find the number of terms it has.
Solution
From the problem above, the first term is 11, that means our (a) = 11
The common difference (d) = 7
So using the nth term formular
Tn = a+(n-1)d
Hope you have not forgotten that?
Here we are told to solve for n (number of terms)
The last term in the problems means that we are giving to equate the value to 144
Tn = a+(n-1)d = 144
Substitute the value for a and d
11 + (n-1)x7 = 144
We then solve the bracket first right? Yes!
11+nx7-1×7=144
11+7n-7=144
Collect like terms
11-7+7n=144
4+7n=144
Take 4 to the right hand side but note the change of sign
7n=144-4
7n=140
Divide both sides by 7
n = 140/7
n = 20
Therefore, the number of terms it has is 20.
Example 4: The 9th term of an AP is 52 while the 12th term is 70. Find the 20th term.
Solution
In this problem, we are to make equation 1 and 2 and then solve it simultaneously
Using the nth term formular
Tn = a+(n-1)d remember n is the …. Of term i.e 7 and 12 respectively
T9 = 52 and T12 = 70
T9 = a+(9-1)d = 52 —– (1)
T12 = a+(12-1)d = 70
a+8d = 52 —– (1)
a+11d = 70 — (2)
Subtract equation 1 from 2 that means
a-a + 11d -8d = 70-52 (2-1)
0 + 3d=18
3d = 18
Divide both sides by the coefficient of d which is 3
d = 18/3
d = 6
Put d = 6 into equation 1 to obtain
a + 8d = 52
a + 8×6= 52
a + 48 = 52
Collect like terms i.e 48 cross the equal sign to the right hand side
a=52 -48
a=4
The 20th term means n=20
T20 = a+(20-1)d
= 4+19×6
4 + 144 = 118
Therefore the 20th term is 118
CONCLUSION:
With this foundation I have laid using these examples, I strongly believe you can strive other challenging problems in nth term of A.P
Make sure you solve the formula because in the subsequent problems involving A.P you are going to need it. Watch out for other topics in mathematics. Trust me it’s amazing, you will love it.
