Chapter 26.7 - General term of a Sequence

In the previous section we saw how to determine the n^th term of a sequence. We also saw a solved example. Now we will see another solved example. Later in this section we will see how this 'discussion about  sequences in general' can be related to Arithmetic progressions.

Solved example 26.37
Fig.26.6 shows a series of squares. The first square have side 1 cm, The second square have side 1¹⁄₂ cm, The third square have side 2 cm, so on. 

Fig.26.6

Write the algebraic expressions for the n^th term of the following sequences: (i) Side (ii) Perimeter (iii) Area (iv) Diagonal
Solution:
Part (i), Side:
1. Side of 1^st square = 1, Side of 2^nd square = 1¹⁄₂, Side of 3^rd square = 2, so on
■ So the sequence is: 1, 1¹⁄₂, 2, 2¹⁄₂, 3, 3¹⁄₂, . . .
⇒1, ³⁄₂, 2, ⁵⁄₂, 3, ⁷⁄₂, . . .⇒¹⁄₁, ³⁄₂, ²⁄₁, ⁵⁄₂, ³⁄₁, ⁷⁄₂, . . .
⇒ ²⁄₂, ³⁄₂, ⁴⁄₂, ⁵⁄₂, ⁶⁄₂, ⁷⁄₂, . . . [Multiplying both numerator and denominator by 2, in those terms which have denominator 1. This will give all terms with denominator 2]
2. Now we see that the numerators form a sequence: 2, 3, 4, 5, . . . 
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [1 more than the 'position number'] of that term
4. So the numerator of the n^th term will be (n+1)
5. So the a_n of the sequence in (1) is given by: a_n = ⁽ⁿ⁺¹⁾⁄₂.

Part (ii), Perimeter:
1. Perimeter of 1^st square = 4×1= 4, Perimeter of 2^nd square = 4×1¹⁄₂ = 6, Perimeter of 3^rd square = 4×2 =8, Perimeter of 4^th square = 4×2¹⁄₂ = 10, so on
■ So the sequence is: 4, 6, 8, 10, . . .
2. Let us write the two tasks and their results:
• First task: To find the rule
Result: It is a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [2 more than twice the 'position number'] of that term
• So the algebraic expression for the n^th term is: a_n = 2n+2 = 2(n+1).

Part (iii), Area:
1. Area of 1^st square = 1 × 1= 1, Area of 2^nd square = 1¹⁄₂ × 1¹⁄₂ = 2.25, Area of 3^rd square = 2 × 2 = 4, Area of 4^th square = 2¹⁄₂ × 2¹⁄₂ = 6.25, Area of 5^th square = 3 × 3 = 9, Area of 6^th square = 3¹⁄₂ × 3¹⁄₂ = 12.25, so on
■ So the sequence is: 1, 2.25, 4, 6.25, . . .
⇒ 1, 2¹⁄₄, 4, 6¹⁄₄, 9, 12¹⁄₄ . . .
⇒1, ⁹⁄₄, 4, ²⁵⁄₄, 9, ⁴⁹⁄₄, . . .⇒¹⁄₁, ⁹⁄₄, ⁴⁄₁, ²⁵⁄₄, ⁹⁄₁, ⁴⁹⁄₄, . . .
⇒ ⁴⁄₄, ⁹⁄₄, ¹⁶⁄₄, ²⁵⁄₄, ³⁶⁄₄, ⁴⁹⁄₄, . . . [Multiplying both numerator and denominator by 4, in those terms which have denominator 1. This will give all terms with denominator 4]
2. Now we see that the numerators form a sequence: 4, 9, 16, 25, 36, 49, . . . 
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form a sequence of squares of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the square of [1 more than the 'position number'] of that term
4. So the numerator of the n^th term will be (n+1)²
5. So the a_n of the sequence in (1) is given by: a_n = ⁽ⁿ⁺¹⁾²⁄₄.

Part (iv), Diagonal:
1. Diagonal of 1^st square = √[(1)² + (1)²] = √[2×(1)²] = √2 × √1 = √2 
Diagonal of 2^nd  square = √[(³⁄₂)² + (³⁄₂)²] = √[2×(³⁄₂)²] = √2 × √[(³⁄₂)²] = √2 × ³⁄₂.
Diagonal of 3^rd square = √[(2)² + (2)²] = √[2×(2)²] = √2 × √[(2)²] = √2 × 2.
Diagonal of 4^th  square = √[(⁵⁄₂)² + (⁵⁄₂)²] = √[2×(⁵⁄₂)²] = √2 × √[(⁵⁄₂)²] = √2 × ⁵⁄₂.
Diagonal of 5^th square = √[(3)² + (3)²] = √[2×(3)²] = √2 × √[(3)²] = √2 × 3.
■ So the sequence is: √2, √2 × ³⁄₂, √2 × 2, √2 × ⁵⁄₂, √2 × 3, . . .
⇒√2 × ²⁄₂, √2 × ³⁄₂, √2 × ⁴⁄₂, √2 × ⁵⁄₂, √2 × ⁶⁄₂, . . .
2. Now we see that the numerators form a sequence: 2, 3, 4, 5, 6, . . . 
3. Let us write the two tasks and their results:
• First task: To find the rule
Result: The numerators form a sequence of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [1 more than the 'position number'] of that term
4. So the numerator of the n^th term will be (n+1)
5. All fractions have denominator 2. Also all fractions are multiplied by √2
So the a_n of the sequence in (1) is given by: a_n =√2 × [⁽ⁿ⁺¹⁾⁄₂]

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• So we now know how to write the algebraic expression for the n^th term of any given sequence. 
• Now we will see the reverse. That is., we will be given  the algebraic expression for the n^th term of the sequence. We must be able to write the sequence. Let us see an example:

Solved example 26.38
Write the first four terms of the sequence whose n^th term is given by: a_n = 2n + 3  
Solution:
We have: a_n = 2n + 3 
Putting n = 1, 2, 3, 4, we get:
a₁ = 2 × 1 + 3 = 2 + 3 = 5,
a₂ = 2 × 2 + 3 = 4 + 3 = 7,
a₃ = 2 × 3 + 3 = 6 + 3 = 9,
a₄ = 2 × 4 + 3 = 8 + 3 = 11
So the required first four terms are 5, 7, 9 and 11

Solved example 26.39
Write the first four terms of the sequence whose n^th term is given by: a_n = n² + 2  
Solution:
We have: a_n = n² + 2  
Putting n = 1, 2, 3, 4, we get:
a₁ = 1 × 1 + 2 = 1 + 2 = 3,
a₂ = 2 × 2 + 2 = 4 + 2 = 6,
a₃ = 3 × 3 + 2 = 9 + 2 = 11,
a₄ = 4 × 4 + 2 = 16 + 2 = 18,
So the required first four terms are 3, 6, 11 and 18

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From the discussion that we had so far, we can make the following conclusion:

■ A sequence can be specified by two methods:
• Method 1: Write sufficient number of terms, starting from the first term a₁. By examining those terms, we will be able to write the algebraic expression for the n^th term a_n
• Method 2: Give the algebraic expression for the n^th term of a sequence. Using that, we will be able to write the first few number of terms

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General term of an Arithmetic progression

We discussed Arithmetic progression in detail earlier in the first section of this chapter. 
■ We saw that, if we know the value of the first term 'a' and the common difference 'd' of an AP, the algebraic expression for the n^th term of that AP can be written as: a_n = a + (n-1)d
• This is the general term of an AP. We may be asked to do different types of problems using this general term. Let us analyse:

In the right side of the algebraic expression a_n = a + (n-1)d, 'a' and 'd' are constants. That is., their values will not change. But 'n' is the 'position number'. It will change.
We can write:
a_n = a + (n-1)d ⇒ a_n =  a + nd - d ⇒ a_n = (a-d) + nd ⇒ a_n = k + nd 
So, if we are given a general term of an A.P, we can straight away identify it's common difference 'd'.
Because, 'd' will be the 'coefficient of n'. Let us see an example:

Solved example 26.40 
Show that the sequence defined by a_n = 3n + 4 is an A.P. Also find it's common difference
Solution:
We have: a_n = 3n + 4
Putting n = 1, 2, 3, 4, we get:
a₁ = 3 × 1 + 4 = 3 + 4 = 7,
a₂ = 3 × 2 + 4 = 6 + 4 = 10,
a₃ = 3 × 3 + 4 = 9 + 4 = 13,
So the first three terms of the sequence are: 7, 10 and 13. We have to prove that they form an A.P. We have done such problems here. Let us use the same steps:
1. Take the last two terms: a_k+1 - a_k = 13 - 10 = 3
2. Take the preceding two terms with one term common: a_k+1 - a_k = 10 - 7 = 3
[In this step we have reached the first term]
3. a_k+1 - a_k is same in the two tests. So it is an AP
4. The common difference d = 3
■ The coefficient of n in the algebraic expression [a_n = 3n + 4] is also 3

Solved example 26.41
Check whether the sequence defined by a_n = 3n² + 2 is an A.P or not
Solution:
We have: a_n = 3n² + 2  
Putting n = 1, 2, 3, we get:
a₁ = 3 × 1 × 1 + 2 = 3 + 2 = 5,
a₂ = 3 × 2 × 2 + 2 = 12 + 2 = 14,
a₃ = 3 × 3 × 3 + 2 = 27 + 2 = 29,
So the first three terms of the sequence are: 5, 14 and 29. We have to check whether they form an A.P. or not. We have done such problems here. Let us use the same steps:
1. Take the last two terms: a_k+1 - a_k = 29 - 14 = 15
2. Take the preceding two terms with one term common: a_k+1 - a_k = 14 - 5 = 9
[In this step we have reached the first term]
3. a_k+1 - a_k is not same in the two tests. So it is not an AP
This can be proved with out using actual values also. The steps are as follows:
1. We have: a_n = 3n² + 2
2. The term after the n^th term, that is., the (n+1)^th term can be obtained by replacing n with (n+1)
So we get: a_(n+1) = 3(n+1)² + 2 = 3 × (n² + 2n +1) + 2 = 3n² + 6n + 3 + 2 = 3n² + 6n + 5
3. The common difference d = a_n+1 - a_n =  3n² + 6n + 5 - (3n² + 2) = 3n² + 6n + 5 - 3n² - 2 = 6n + 3
4. (6n+3) is not independent of n. That means it will vary with the position of the term. So it is not an A.P
■ In general, if the algebraic expression for the n^th term is a linear expression in n, it will be an A.P

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We have completed the present discussion on Arithmetic progressions. In the next chapter, we will see Angles inside circles.

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Chapter 26.6 - Basic details about Sequences

In the previous section we completed the discussion on Arithmetic progressions. In this section we will see sequences in general.

• Consider the following arrangement of numbers:
1, 4, 9, 16, . . .
• In this arrangement, the numbers are arranged according to a rule. We must do the following two tasks:
1. Find the rule
2. Find how the rule is applied

1. Finding the rule:
• If we carefully examine the numbers, the rule can be found out without much difficulty. We find that the numbers are all squares.
    ♦ 1 is the square of 1 
    ♦ 4 is the square of 2
    ♦ 9 is the square of 3
so on
• So the rule is: Squares of natural numbers are arranged in a sequence
2. Finding how the rule is applied:
• For this task, we have to first familiarise ourselves with some general procedures associated with sequences. Let us see them:
(a) The various numbers occurring in the sequence are called terms
(b) The terms are denoted as a₁, a₂, a₃, . . . etc., OR x₁, x₂, x₃, . . . etc.,
(c) The subscripts denote the position of the terms. 
    ♦ a₁ is the 1^st term
    ♦ a₂ is the 2^nd term
    ♦ a₃ is the 3^rd  term
so on
■ So subscripts are the 'position numbers'
Now let us proceed with our second task:
• For our given sequence, we can write:
a₁ = 1², a₂ = 2², a₃ = 3² so on
■ From the above step, we can note one point:
Consider any position. The actual term at that position will be the square of the 'position number'. For example:
    ♦ a₅ will be 5².
    ♦ a₉ will be 9².
■ So, if we consider the n^th term,
• It is denoted as a_n
• It's 'position number' is n
• The actual term at that n^th position is n².
■ So we can write: a_n = n².
Thus our second task is also complete. Let us summarise:
• First task: To find the rule
Result: It is a sequence of squares of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the square of the 'position number' of that term. 
• This gives an expression: a_n = n²
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.

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■ To understand the two tasks and their results, let us consider a variant of the above sequence:

0, 1, 4, 9, 16, . . .
The tasks and their results are:
• First task: To find the rule
Result: It is a sequence of squares of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the square of [1 subtracted from the 'position number'] of that term
• This gives an expression: a_n = (n-1)² 
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.
• Note that, in the variant, rule is the same as before. But the way in which the rule is applied is different
■ Another variant:
1, 0, 1, 4, 9, 16, . . .
The tasks and their results are:
• First task: To find the rule
Result: It is a sequence of squares of natural numbers
• Second task: To find how the rule is applied
• Result: Any term taken from the sequence is the square of [2 subtracted from the 'position number'] of that term
• This gives an expression: a_n = (n-2)² 
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.
• Note that, in this variant also, rule is the same as before. But the way in which the rule is applied is different 

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■ Now we will see another example. Consider the sequence:

1, ¹⁄₂, ¹⁄₃, ¹⁄₄, ¹⁄₅, . . .
In this sequence, a₁ = ¹⁄₁, a₂ = ¹⁄₂, a₃ = ¹⁄₃, so on  
Let us write the two tasks and their results:
• First task: To find the rule
Result: It is a sequence of reciprocals of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the reciprocal of the 'position number' of that term
• This gives an expression: a_n = ¹⁄_n
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.
■ One more example:
1, 3, 5, 7, 9, . . .
In this sequence, a₁ =1, a₂ = 3, a₃ = 5, so on  
Let us write the two tasks and their results:
• First task: To find the rule
Result: It is a sequence of odd numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is [1 subtracted from twice the 'position number'] of that term
• This gives an expression: a_n = 2n - 1
    ♦ This expression shows how 'the actual value of any given term' is related to the 'position number' of that term.

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So we saw three examples of sequences. In each of them, we performed the two tasks and obtained their results. We can now give the definition for sequences:
■ A sequence is an arrangement of numbers in a definite order according to some rule

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In any sequence, as a result of performing the two tasks, we will get an expression for a_n, which is the n^th term.
• In the first example we got: a_n = n²   
• In the second example we got: a_n = ¹⁄_n
• In the third example we got: a_n = 2n - 1 
■ This n^th term is called the general term of the sequence.

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We will now see some solved examples:
Solved example 26.36
One cubic cm of iron weighs 7.8 grams. Write as sequence:
(i) The volumes of iron cubes of sides 1 cm, 2 cm, 3 cm, . . .
(ii) The weights of iron cubes of sides 1 cm, 2 cm, 3 cm, . . .
(iii) Also write the algebraic expression for the n^th term of the above sequences
Solution:
Part (i), Volumes:
• Volume of a cube of side 1 cm = 1³ = 1 cm³
• Volume of a cube of side 2 cm = 2³ = 8 cm³
• Volume of a cube of side 3 cm = 3³ = 27 cm³.
■ So the sequence is: 1, 8, 27, . . .
Part (ii), Weights:
• Weight of a cube of side 1 cm = volume × weight per unit volume = 1 × 7.8 = 7.8 grams
• Weight of a cube of side 2 cm = 8 × 7.8 = 62.4 grams
• Weight of a cube of side 3 cm = 27 × 7.8 = 210.6 grams
■ So the sequence is: 7.8, 62.4, 210.6, . . .
Part (iii), Algebraic expressions
1. Consider the first sequence:
1, 8, 27, . . .
In this sequence, a₁ =  1³ = 1 , a₂ = 2³ = 8, a₃ = 3³ = 27, so on  
• First task: To find the rule
Result: It is a sequence of cubes of natural numbers
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the cube of the 'position number' of that term. 
• So the algebraic expression for the n^th term is: a_n = n³.
2. Consider the second sequence:
7.8, 62.4, 210.6, . . .
In this sequence, a₁ = 7.8 × 1³ = 7.8, a₂ = 7.8 × 2³ = 62.4, a₃ = 7.8 × 3³ = 210.4, so on  
• First task: To find the rule
Result: It is a sequence of cubes of natural numbers multiplied by 7.8
• Second task: To find how the rule is applied
Result: Any term taken from the sequence is the cube of the 'position number' of that term, multiplied by 7.8  
• So the algebraic expression for the n^th term is: a_n = 7.8n³

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In the next section we will see another solved example. We will also see how the above discussion is related to Arithmetic progressions.

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Chapter 26.5 - Sum of n terms - Solved examples

In the previous section we completed the discussion on how to determine the sum of first n terms of an AP. We also saw some solved examples. In this section we will see a few more solved examples.

Solved example 26.28
Find the sum of the following APs
(i) 2, 7, 12, . . . , to 10 terms
Solution:
1. In the given AP, a = 2, d = 12 -7 = 5 and n = 10
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] ⇒ S = ¹⁰⁄₂ × [2 × 2 + (10-1)×5]
= 5 × [4 + 45] = 5 × [49] = 5 × 49 = 245

(ii)  ¹⁄₁₅, ¹⁄₁₂, ¹⁄₁₀,, . . . , to 11 terms
Solution:
1. In the given AP, a = ¹⁄₁₅, d = ¹⁄₁₀ - ¹⁄₁₂ = ¹⁄₆₀ and n = 11
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] ⇒ S = ¹¹⁄₂ × [2 × ¹⁄₁₅ + (11-1)×¹⁄₆₀]
= ¹¹⁄₂ × [²⁄₁₅ + 10 × ¹⁄₆₀] = ¹¹⁄₂ × [²⁄₁₅ + ¹⁄₆] = ¹¹⁄₂ × [¹⁸⁄₆₀] = ¹¹⁄₂ × [³⁄₁₀] = ³³⁄₂₀

Solved example 26.29
How many terms of the AP: 24, 21, 18, . . . must be taken so that, their sum is 78?
Solution:
1. In the given AP, a = 24, d = 18 -21 = -3 and S_n = 78
2. We have to find n
3. We have: S_n = ⁿ⁄₂ [2a + (n-1)d] ⇒ 78 = ⁿ⁄₂[2 × 24 + (n-1)×-3] ⇒ 78 = ⁿ⁄₂[48 - 3n + 3]
⇒ 78 = ⁿ⁄₂[51 - 3n] ⇒ 156 = n[51 - 3n] ⇒ 156 = 51n - 3n² ⇒ 3n² - 51n + 156 = 0
4. Solving this equation, we get n = 4 and n = 13
5. So, sum of the first 4 terms of the AP starting from 24 will be equal to 78
6. Sum of the first 13 terms starting from 24 will also be equal to 78
7. How can the two sums be equal to 78?
Ans: (i) Consider the terms starting from the fifth upto the thirteenth
(ii) The sum of those terms will be zero. Because, some of those terms are positive, and the others negative. Those terms will cancel out each other.
This is shown in a tabular form here.

Solved example 26.30
Find the sums given below:
(i) 7 + 10¹⁄₂ + 14 + . . . + 84
Solution:
1. The given series is an AP because, 14 - 10¹⁄₂ = 3¹⁄₂.
Also 10¹⁄₂ - 7 = 3¹⁄₂.
2. So, in the given AP, a = 7, d = 3¹⁄₂ and last term l = 84
3. To find the sum, we need to know the number of terms. That means, we need to know the 'n value' of the last term 84. We did such problems in a previous section. [See solved example 26.15 at the beginning of section 26.3]
(i) We have:  n^th term = a + (n-1)d ⇒ 84 = 7 + (n-1) × 3¹⁄₂ 
⇒  84 = 7 + ⁷ⁿ⁄₂ - ⁷⁄₂ ⇒ ⁷ⁿ⁄₂ = 84 - 7 + ⁷⁄₂ ⇒ ⁷ⁿ⁄₂ = ¹⁶¹⁄₂ ⇒ 7n = 161 ⇒ n = ¹⁶¹⁄₇ = 23
(ii) So 84 is the 23^rd term of the given AP.
4. Now we know the first term, last term and n. We can use eq.26.2(b) that we saw in the previous section.
S = ⁿ⁄₂ [a + l] ⇒ S = ²³⁄₂ [7 + 84] = ²³⁄₂ [91] = 1046¹⁄₂.

(ii) -5 + (-8) + (-11) + . . . + (-230)
Solution:
1. The given series is an AP because, -11 - (-8) = -11 + 8 = -3.
Also -8 - (-5) = -8 + 5 = -3.
2. So, in the given AP, a = -5, d = -3 and last term l = -230
3. To find the sum, we need to know the number of terms. That means, we need to know the 'n value' of the last term -230. We did such problems in a previous section. [See solved example 26.15 at the beginning of section 26.3]. Also see the previous problem.
(i) We have:  n^th term = a + (n-1)d ⇒ -230 = -5 + (n-1) × -3 ⇒ -230 = -5 -3n + 3 
⇒ 3n = 230 -5 + 3 ⇒ 3n = 228 ⇒ n = 76
(ii) So -230 is the 76^th term of the given AP.
4. Now we know the first term, last term and n. We can use eq.26.2(b) that we saw in the previous section.
S = ⁿ⁄₂ [a + l] ⇒ S = ⁷⁶⁄₂ [-5 + -230] = ⁷⁶⁄₂ [-235] = -8930 

Solved example 26.31
In an AP: Given a = 5, d = 3, a_n = 50, find n and S_n.
Solution:
1. In this problem, the n^th term a_n is given as 50. 
2. The sum from the first term 5 to the n^th term 50 is to be calculated. 50 is the last term.
3. To find the sum, we need to know the number of terms. That means, we need to know the 'n value' of the last term 50. We did such problems in a previous section. [See solved example 26.15 at the beginning of section 26.3]. Also see the previous problem. 
(i) We have:  n^th term = a + (n-1)d ⇒ 50 = 5 + (n-1) × 3 ⇒ 50 = 5 +3n - 3 
⇒ 3n = 50 -5 + 3 ⇒ 3n = 48 ⇒ n = 16
(ii) So 50 is the 16^th term of the given AP.
4. Now we know the first term, last term and n. We can use eq.26.2(b) that we saw in the previous section.
S = ⁿ⁄₂ [a + l] ⇒ S = ¹⁶⁄₂ [5 + 50] = 8 × 55 = 440

Solved example 26.32
Given a = 8, a_n  = 62, S_n = 210, find n and d.  
Solution:
1. We know the first term, last term and S_n. We can use eq.26.2(b) that we saw in the previous section.
S = ⁿ⁄₂ [a + l] ⇒ 210 = ⁿ⁄₂ [8 + 62] ⇒ 210 = ⁿ⁄₂ [70] ⇒ n = ²¹⁰⁄₃₅ = 6
• So the 6^th term is 62
2. We can use eq.26.1 which gives the position of any term:
• n^th term = a + (n-1)d ⇒ 62 = 8 + (6-1)d ⇒ 62 = 8 + 5d ⇒ 5d = 54 ⇒ d = ⁵⁴⁄₅.

Solved example 26.33
Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
Solution:
1. S₁₀, which is the sum of first 10 terms is given as 125
• The first term a is not given. So we will use the basic eq.26.2 which gives the sum of n terms:
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] ⇒125 = ¹⁰⁄₂ × [2a + (10-1)×d] 
⇒125 = 5 × [2a + 9d] ⇒25 = 2a + 9d
3. The third term a3 is given as 15. We can use eq.26.1 which gives the position of any term:
• n^th term = a + (n-1)d ⇒ 15 = a + (3-1)d ⇒ 15 = a + 2d
4. So we have two equations:
(i) 2a + 9d = 25 
(ii) a + 2d = 15
• From (ii) we get: a = 15-2d
• Substituting this in (i) we get: 2 × (15-2d) + 9d = 25 ⇒ 30 -4d +9d = 25 
⇒ 30 + 5d = 25 ⇒ 5d = -5 ⇒ d = -1
• Substituting this value of d in (ii) we get: a + 2 ×-1 = 15 ⇒ a -2 = 15 ⇒ a = 17  
5. Now we can find a₁₀. We have:
• n^th term = a + (n-1)d ⇒ a₁₀ = 17 + (10-1)×-1= 17 - 9 = 8
The AP can be seen here in a tabular form

Solved example 26.34
A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find : (i) the production in the 1^st year (ii) the production in the 10^th year (iii) the total production in first 7 years
Solution:
• The production increases uniformly by a fixed number every year. So it is an AP
• We are given: a₃ = 600 and a₇ = 700
Part (i): We have to find the first term a
1. n^th term = a + (n-1)d ⇒ a₃ = 600 = a + (3-1)d ⇒ 600 = a + 2d
2. Again, n^th term = a + (n-1)d ⇒ a₇ = 700 = a + (7-1)d ⇒ 700 = a + 6d 
3. Now we have to solve the two equations.
(i) From (1) we have: a = 600-2d
(ii) Substituting this value of a in (2) we get:
700 = 600 - 2d + 6d ⇒ 700 = 600 + 4d ⇒ 4d = 100 ⇒ d = 25
(iii) substituting this value of d in (1) we get:
600 = a + 2 × 25 ⇒ 600 = a + 50 ⇒ a = 550
Part (ii): We have to find a₁₀
1. n^th term = a + (n-1)d ⇒ a₁₀ = 550 + (10-1)×25= 550 + 9 × 25 = 550 + 225 = 775
Part (iii): We have to find S₇
1. We have: S = ⁿ⁄₂ [2a + (n-1)d] = ⁷⁄₂ × [2 × 550 + (7-1)×25] 
= ⁷⁄₂ × [1100 + 6×25] = ⁷⁄₂ × [1100 +150] = ⁷⁄₂ × 1250 = 7 × 625 = 4375

Solved example 26.35
A contract on construction job specifies a 'penalty for delay of completion' beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
Problem analysis:
■ The contractor has to complete a construction job on a specified date. Let this date be March 31^st.
• If the contractor completes the work only on April 1^st, it means that he has delayed the work by 1 day
    ♦ For this delay of 1 day, he has to pay Rs 200 
• If the contractor completes the work only on April 2^nd, it means that he has delayed the work by 2 days
    ♦ For this delay of  2 days, he has to pay Rs 200 + 250
• If the contractor completes the work only on April 3^rd, it means that he has delayed the work by 3 days
    ♦ For this delay of 3 days, he has to pay Rs 200 + 250 + 300
So on . . .
■ Thus we find that, it is an AP with a = 200 and d = 50
1. For a delay of 30 days, there will be 30 terms. So n = 30
2. We have: S = ⁿ⁄₂ [2a + (n-1)d] = ³⁰⁄₂ × [2 × 200 + (30-1)×50] 
= 15 × [400 + 29×50] = 15 × [400 +1450] = 15 × 1850 = Rs 27750

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In the next section we will see Sequences in general.

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