Progressions

Arithmetic Progression

Quantities said to be in Arithmetic progression when they increase or decreases by a common difference.
For example :
4, 6, 8, 10 ........ (Increasing by common difference 2.)
8, 2, -4, -10 ........... (decreasing by common difference 6.)
A general series of an AP is given by,
a, a+d, a+2d, a+3d, a+4d ...........(Here common difference is d.)

Common difference :

The common difference of an AP is found by subtracting any term of the series from next term.
That is, common difference of an AP = (Tn - Tn-1).
For the 1st above example common difference
= 6 - 4 = 8 - 6 = 10 - 8 = 2
For the 2nd above example common difference
= 2 - 8 = -4 - 2 = -10-(-4) = -6
And for 3rd example common difference = d

The rth term of an AP:

If we examine the above series a, a+d, a+2d, a+3d, a+4d ......... , we notice that in any term the coefficient of d is always less by one than the position of that term in the series.
Thus, rth term of an AP is given by, Tr = a +(r-1)d.

Last term of an AP:

If n be the number of terms, and if L denotes the last term or the nth term, we have
L = a + (n-1)d.
Or, Tn = a + (n-1)d.

Sum of an AP:

If we examine the series a, a+d, a+2d, a+3d, a+4d ..........
Here, the first term is a and common difference is d . Let it has n total number of terms. Also, let L denote the last term and S be the required sum; then
S = $$n \times \frac{{a + L}}{2} = {\text{No}}{\text{.}}\,{\text{of}}\,{\text{terms}} \times $$     $$\frac{{{\text{first}}\,{\text{term}} + {\text{last}}\,{\text{term}}}}{2}$$
S = $$\frac{n}{2} \times \left[ {2a + \left( {n - 1} \right) \times d} \right];$$     [by putting L = a + (n-1) × d, in the above equation).

Arithmetic Mean:

Let A be the Arithmetic mean of two quantities a and b. Then since a, A, b are in AP. Then,
b - A = A - a;
Or, A = $$\frac{{{\text{a}} + {\text{b}}}}{2}$$

Process for finding the nth term of an AP

Suppose we have to find the 17th term of the AP 3, 7, 11 .....
It could be given by formula,
Tn = a + (n-1) × d.
T17 = 3 + (17-1) × 4 = 3 + 16 × 4 = 67

Mechanically, if we insert the value for a, n and d and we will get this answer.
However, if we replace the above process with a mental process, then we get answer faster. It is as follow:
In order to find the 17th term of above sequence add the common difference to the first term, sixteen times.

i.e. 17th term = a + 16 times of common difference.
17th term = 3 + 4 × 16 = 67. (Sixteen, since it is one less than 17)

Similarly, for finding 37th term of AP 3, 7, 11 ....... Could be given by
37th term = 3 + 4 × 36 = 3 + 288 = 291.

Average of an AP and Corresponding terms of the AP:

Let us consider an AP 2, 6, 10, 14, 18, 22.
Average of this AP is given by, $$\frac{{2 + 6 + 10 + 14 + 18 + 22}}{6} = 12.$$
12 is also the average of 1st term and last term, $$\frac{{2 + 22}}{2} = 12;$$
12 is also the average of 2nd term and 5th term, $$\frac{{6 + 18}}{2} = 12;$$
In fact, 12 is also the average of 3rd term and 4th term, $$\frac{{10 + 14}}{2} = 12.$$

The pairs of term of an AP which gives the same average as the original AP has, are called Corresponding Terms in the AP.

If we try to notice the sum of the numbers of the pair of corresponding terms, we get,
1st and 6th (so that 1 + 6 = 7)
2nd and 5th (so that 2 + 5 = 7)
3rd and 4th (so that 3 + 4 = 7)

In each case, the sum of the term numbers for the terms in a corresponding pair is one greater than the number of terms of the AP. And this holds true for all AP.

For example:
If an AP has 34 terms then for instance, we can predict that the 7th will have the 28th term as its corresponding term, or for 9th term 26th will be the corresponding term (since 9 + 26 = 35).

Short-Cut Process for finding Sum of an AP:

Once we get the corresponding terms for any AP, we can easily find the sum of an AP by using property of averages:

Sum = Number of terms × Average of that AP.
This process is much time saving than conventional formula,
Sum = $$\frac{{\text{n}}}{2} \times \left[ {2a + \left( {n - 1} \right) \times d} \right];$$

For example:
For an AP, 2, 6, 10, 14, 18, 22.
Average = $$\frac{{2 + 22}}{2} = 12;$$
Sum = number of terms × average
Sum = 6 × 12 = 72

Short-cut Process to find common difference of an AP if two terms of AP given

If 3rd term of an AP given as 8 and 8th term given as 28. We can visualize as,
_ _ 8 _ _ _ _ 28.

We see that 3rd term of the AP is moving 8 to 28 to give 8th term.
(8 to 28) net addition to third term is 20. We can see that there is five times of addition of common difference to 8. Then common difference = $$\frac{{20}}{5}$$ = 4.

Illustrations:
Find the sum of an AP of 17 terms, whose 3rd term is 8 and 8th term is 28.
Since, 8th is 28 and d =4, the 10th term becomes 28 + 4 + 4 = 36.
Average of the AP = average of 8th term and 10th term = $$\frac{{28 + 36}}{2}$$  = 32;
Hence, Sum = number of terms × average;
Sum = 17 × 32 = 544.

Key Facts:

If same quantity added to or subtracted from all the term terms of an AP, the resulting terms will form an AP, but the same common difference as before.
E.g. 4, 8, 12, 16 ...... (d = 4). If we add 2 to the all the terms of above AP, then 6, 20, 14, 18 ....... (d = 4).

If all the terms of an AP be multiplied or divided by the same quantity, the resulting terms will form an AP, but with a new common difference, which will be multiplication or division of old common difference.
E.g. 4, 8, 12, 16 ..... (d = 4). On multiplying it by 2. 8, 16, 24, 32 ....... (d = 4 × 2 = 8).

If you have to assume 3 terms in AP, assume as a-d, a, a+d or a, a+d, a+2d.

Sum of first n natural numbers is given by,
S = 1 + 2 + 3 + ........ + n,
S = $$\frac{{{\text{n}} \times \left( {{\text{n}} + 1} \right)}}{2}.$$

Sum of the squares of the first n natural numbers is given by,
S = 12 + 22 + 32 + ........ + n2
S = $$\frac{{{\text{n}} \times \left( {{\text{n}} + 1} \right)\left( {2{\text{n}} + 1} \right)}}{6}.$$

Sum of the cubes of the first n natural numbers is given by,
S = 13 + 23 + 33 + ......... + n3
S = $${\left[ {\frac{{{\text{n}} \times \left( {{\text{n}} + 1} \right)}}{2}} \right]^2}$$
S = [Sum of n natural numbers]2
Thus, the sum of the cubes of the first n natural numbers is equal to the square of the sum of these numbers.

Sum of the first n odd natural numbers is given by,
S = 1 + 3 + 5 + ........ +(2n-1)
S = n2.

Sum of first n even natural numbers is given by,
S = 2 + 4 + 6 + ....... + 2n.
S = n × (n+1)
S= n2 + n

Sum of odd numbers = n, where n is a natural number,
case 1: If n is odd ==> $${\left[ {\frac{{{\text{n}} + 1}}{2}} \right]^2}$$
case 2: If n is even ==> $${\left[ {\frac{{\text{n}}}{2}} \right]^2}$$

Sum of even numbers = n, where n is a natural number,
case 1: If n is odd ==> $$\left\{ {\frac{{\text{n}}}{2} \times \left[ {\frac{{\text{n}}}{2} + 1} \right]} \right\}$$
case 2: If n is even ==> $$\left[ {\frac{{{\text{n}} - 1}}{2}} \right] \times \left[ {\frac{{{\text{n}} + 1}}{2}} \right]$$

Geometric Progression:

Quantities are said to be in Geometric Progression when they increases or decreases by a constant factor.
This constant factor is also called the common ratio and it is found by dividing any the term immediately preceding it.
General series of a GP is given as,
a, ar, ar2, ar3, .........

It is noticeable that in any term the index of r is always less by one than the number of term in the series.
Last term or nth term of GP = ar(n-1).
Let a and b in GP, G is geometric mean then since,
A, G, b are in GP,
$$\frac{{\text{b}}}{{\text{G}}}$$ = $$\frac{{\text{G}}}{{\text{a}}}$$; Or, G2 = ab
Hence, G = vab.

Sum of GP

If r > 1 then,
Sum (Sn) = $${\text{a}} \times \frac{{{{\text{r}}^{\text{n}}} - 1}}{{{\text{r}} - 1}};$$
If r < 1, then,
Sum (Sn) = $$a \times \frac{{1 - {{\text{r}}^{\text{n}}}}}{{1 - {\text{r}}}};$$
For a GP which has infinite number of terms,
Sum (Sn) = $$\frac{{\text{a}}}{{1 - {\text{r}}}};$$

Click Here for Solved Examples on AP and GP

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Comments ( 1 )

  1. voora anand
    7 years ago
    super