Progressions Complete Tutorial - I - CAT 2009 Online Preparation

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CAT 2009 Online Preparation

Progressions Complete Tutorial - I

Author: Sureshbala

In this chapter, we will look at Arithmetic Progression and Geometric progression. While there is a third type of progression-Harmonic Progression - its usefulness from the point of view of the competitive examinations is very low and hence we will not be examinations is very low and hence we will not be covering it here. Knowledge of progressions will also help in answering questions related to number series.

ARITHMETIC PROGRESSION (A.P.)

Quantities are said to be in Arithmetic Progression when they increase by a constant difference to get the next or the previous term respectively.

An A.P can be represented by a, a + d, a + 2d ….. [a + (n - 1) d]. Here, quantity d is added to any term to get the next term of the progression.

a is called the first term of the progression, n is the number of terms in the progression; d is called the common difference. The $n^{th}$ term is normally represented by T_n and the sum to terms of an A.P. is denoted by S_n

$T_n = n^{th}$ terms = a + (n - 1)d

$S_n = \text{sum of n terms}$ $=\cfrac{n}{2} \times [2a + (n - 1)d]$

The sum to terms of an A.P. can also be written in a different manner.

sum of terms $= \cfrac{n}{2} \times [2a + (n-1)d]$

$= \cfrac{n}{2} \times [a + \{a + (n - 1) d \}]$

But, where there are n terms in an A.P. a is the first term and {a+(n-1)d} is the last term. Hence,

$S_n = \cfrac{n}{2} \times [\text{First Term + Last Term}]$

The average of all terms in an A.P. is called their Arithmetic Mean (A.M). since average is equal to the { sum of all the quantities/number of quantities}. A.M must be equal to the sum of the A.P divided by the number of terms in the A.P.

Arithmetic Mean of n terms in A.P.

$\cfrac{S_n}{n}= \cfrac{1}{2} \{2a + (n - 1) d \}$
$=\cfrac{1}{n} \times \cfrac{n}{2} \times (Fist \; Term + Last \; Term)$
$= \cfrac{(Fist \; Term + Last \; Term)}{2}$

i.e., A.M. is the average of the first and the last terms of the A.P.

Arithmetic Mean can also be obtained by considering any two terms which are EQUIDISTANT from both ends of the A.P. and taking their average , i.e.,

- the average of the second term from the beginning and the second term from the end will be equal to the A.M.

- the average of the third term from the beginning and the third term from the end will also be equal to the A.M. and so on.

In general, the average of the $K^{th}$ term from the beginning and the $K^{th}$ term from the end will be equal to the A.M.

Conversely, if the A.M. of an A.P. is known, the sum to n terms of the series (S_n) can be expressed as $S_n = n \times A.M.$

If three numbers are in arithmetical progression the middle number is called the Arithmetic mean, i.e. if a, b , c are in A.P., then b is the A.M. of the three terms and $b = \cfrac{a+c}{2}$

If a and b are in Arithmetic Progressions (A.P.) then there A.M $= \cfrac{a+b}{2}$

If three numbers are in A.P., we can take the four terms to be (a-d), a and (a + d)

If four numbers are in A.P., we can take the four terms to be (a-3d), (a - d) (a + d) and (a+3d)

If five numbers are in A.P., we can take the five terms to be (a - 2d), (a - d), a, (a + d) and (a+2d)

EXAMPLES

1. Find the $14^{th}$ term of an A.P., whose first term is 3 and common difference is 2.

Sol: $n^{th} term - a + (n-1)d$ , where a is the $1^{st}$ term, d is the common difference.

$\therefore 14^{th} term = 3+ (14-1)2 = 29$

2. Find the number of terms in an A.P., if the first term is 2 and the last term is 41. Given, the common difference is 3.

Sol: Last term in A.P = a+(n-1)d = 2+(n-1)3

$\Rightarrow 3n -1 = 41$
$3n = 42 \Rightarrow n = 14$

So, there are 14 terms in the A.P.

3. Find the first term and the common difference of an A.P, if the $3^{rd}$ term is 6 and the $17^{th}$ term is 34.

Sol: If a is the first term and the common difference d, then we have

a + 2d = 6 …… (1)
a + 16d = 34…… (2)

(2) - (1) gives
$14d = 28 \Rightarrow d = 2$

Substituting this values of d in (1), we get a= 2

$\therefore a = 2 \; and \; d = 2$

4. Find the $1^{st}$ term, common difference and number of terms of an A.P given its $4^{th}$ term is 12, $16^{th}$ term is 20 and last term is 26.

Sol: Let $1^{st} term = a$

Common difference = d

Number of terms = n
a + 3d = 12 ….. (1)

a + 15d = 20 …. (2)

a + (n-1)d =26 ………(3)

Solving (1) and (2) we get d = 2/3 and a = 10

Substituting these values in (3) we get n =25.

5. Find the sum of the first 22 terms of an A.P given the first term is 4 and the common difference is 4/3.

Sol: Sum of 22 terms

$= \cfrac{22}{2} \times [(2 \times 4) + (22-1) \times \cfrac{4}{3}] = 11 \times (8+28) = 396$

6. Find the arithmetic mean of the A.P with 41 terms, whose first term is 2.5 and common difference is 0.75

Sol:
$\text{Arithmetic mean} = \cfrac{1}{2}[2a + (n-1)d]$
=1/2[2(2.5) + (41-1)0.75] = 17.5

7. Divide 124 into four parts which are in A.P such that the product of the $1^{st} \And 4^{th}$ part is 128 less than the product of the $2^{nd} \And 3^{rd}$ parts.

Sol: Let the 4 parts be (a - 3d), (a - d), (a + d) and (a + 3d). The sum of these 4 parts is 124,

$i.e. , \Rightarrow 4a = 124 \Rightarrow a = 31$
(a-3d)(a+3d) = (a-d)(a+d) - 128

$\Rightarrow a^2 -9d^2 = a^2 - d^2 - 128$
$\Rightarrow 8d^2 = 128 \Rightarrow d = \overset{+}{-} 4$

As we know a = 31

the 4 parts are 19, 27, 35, 43

Note: When value of d is taken as -4, the same four numbers are obtained, but in decreasing order.

8. Find the three terms in A.P., whose sum is 36 and product is 960.

Sol: Let the three terms be (a - d), a and (a + d).

Sum of these terms is 3a.

Product of these three terms is

$(a+d)a \; (a-d) = 960 \Rightarrow (12+d)(12-d)=80$ $\Rightarrow 144 - d^2 = 80 \Rightarrow d = \overset{-}{+}8$

If d = 8, the terms are 4, 12, 20

Note: when d =-8 is considered, the same values of the numbers are obtained.

9. The first term of an A.P., is 7 and the last term is 47. If its sum is 297, find the number of terms and the common difference.

Sol: if number of terms is n, then we have

$\cfrac{n}{2} = (7+47) = 297 \Rightarrow n =11$

If d is the common difference $11^{th}$ term

$= 7+(10)d = 47 \Rightarrow d =4$

10. The $63^{th} \; \text{and} \; 6^{th}$ terms of an A.P are -77 and 37. Find the $17^{th}$ term.

Sol: a + 62d= - 77

$a + 5d = 37 \Rightarrow d = -2 \And a = 47$
$\therefore 17^{th} term = a + 16d = 15$

11. The sum of three numbers in A.P is 21 and the sum of their squares is 165. Find the three numbers.

Sol: Let the 3 numbers be a - d, a and a + d

$(a-d) + a+(a+d) = 21 \Rightarrow a = 7$
$(a-d)^2 + a^2 + (a+d)^{2} = 165$
$\Rightarrow 3a^2 + 2d^2 = 165 \Rightarrow 2d^2 = 165 - 3(7)^2$
$\Rightarrow 2d^2 18 \therefore d = \overset{+}{-} 3$

Hence the 3 numbers are 4, 7 and 10.

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