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Question

Question: Calculate the sum of the sequences given below: \((a)\;7 + 10\dfrac{1}{2} + 14 + ... + 84\) \((b...

Calculate the sum of the sequences given below:
(a)  7+1012+14+...+84(a)\;7 + 10\dfrac{1}{2} + 14 + ... + 84
(b)  34+32+30+...+10(b)\;34 + 32 + 30 + ... + 10
(c)  5+(8)+(11)+...+(230)(c)\; - 5 + ( - 8) + ( - 11) + ... + ( - 230)

Explanation

Solution

Each of these subdivisions can be solved in the same manner. We can see that each of these questions show a similar pattern. The pattern the series are showing is that they are having a common difference between consecutive terms. Common difference means that the series shows an arithmetic progression. Remember that the formula for sum of nnterms of a series is:
Sn=n2[2×a+(n1)×d]\Rightarrow Sn = \dfrac{n}{2}[2 \times a + (n - 1) \times d], here a=a = first term in the sequence, d=d = common difference between terms, n=n = number of terms in the sequence

Complete step-by-step solution:
(a)(a) First let us check what kind of pattern the series follows:
First three terms
a1=7,  a2=1012,  a3=14a_1 = 7,\;a_2 = 10\dfrac{1}{2},\;a_3 = 14
Checking the difference between consecutive terms:
a2a1=2127=3.5 a3a2=14212=3.5  \Rightarrow a_2 - a_1 = \dfrac{{21}}{2} - 7 = 3.5 \\\ \Rightarrow a_3 - a_2 = 14 - \dfrac{{21}}{2} = 3.5 \\\
So there is a common difference, let it be d=3.5d = 3.5
To find number of terms nn, we can use the formula;
an=a+(n1)×dan = a + (n - 1) \times d
84=7+(n1)×3.5\Rightarrow 84 = 7 + (n - 1) \times 3.5
n=23\Rightarrow n = 23
Now clearly sum of n=23 \Rightarrow n = 23 terms will be;
Sn=232[2×7+(231)×3.5]\Rightarrow Sn = \dfrac{{23}}{2}[2 \times 7 + (23 - 1) \times 3.5]
S23=104612\Rightarrow S23 = 1046\dfrac{1}{2}
So the sum of this sequence is: 1046121046\dfrac{1}{2}

(b)(b) First let us check what kind of pattern the series follows:
First three terms
a1=34,  a2=32,  a3=30a_1 = 34,\;a_2 = 32,\;a_3 = 30
Checking the difference between consecutive terms:
a2a1=3432=2 a3a2=3032=2  \Rightarrow a_2 - a_1 = 34 - 32 = - 2 \\\ \Rightarrow a_3 - a_2 = 30 - 32 = - 2 \\\
So there is a common difference, let it be d=2d = - 2
To find number of terms nn, we can use the formula;
an=a+(n1)×dan = a + (n - 1) \times d
10=34+(n1)×2\Rightarrow 10 = 34 + (n - 1) \times - 2
n=13\Rightarrow n = 13
Now clearly sum of n=13 \Rightarrow n = 13 terms will be;
Sn=132[2×34+(131)×2]\Rightarrow Sn = \dfrac{{13}}{2}[2 \times 34 + (13 - 1) \times - 2]
S13=286\Rightarrow S13 = 286
So the sum of this sequence is: 286286

(c)(c) First let us check what kind of pattern the series follows:
First three terms
a1=5,  a2=8,  a3=11a_1 = - 5,\;a_2 = - 8,\;a_3 = - 11
Checking the difference between consecutive terms:
a2a1=8+5=3 a3a2=11+8=3  \Rightarrow a_2 - a_1 = - 8 + 5 = - 3 \\\ \Rightarrow a_3 - a_2 = - 11 + 8 = - 3 \\\
So there is a common difference, let it be d=3d = - 3
To find number of terms nn, we can use the formula;
an=a+(n1)×dan = a + (n - 1) \times d
230=5+(n1)×3\Rightarrow - 230 = - 5 + (n - 1) \times - 3
n=76\Rightarrow n = 76
Now clearly sum of n=76 \Rightarrow n = 76 terms will be;
Sn=762[2×5+(761)×3]\Rightarrow Sn = \dfrac{{76}}{2}[2 \times - 5 + (76 - 1) \times - 3]
S23=8930\Rightarrow S23 = - 8930
So the sum of this sequence is: 8930 - 8930

Note: The division of the numerator by the denominator can be a cumbersome task as it involves thorough knowledge of algebraic rules and long division method. Care should be taken while doing the same and proceeding with the process to convert a fraction into decimal. The decimal expansion of a rational number in pq\dfrac{p}{q} form can be terminating or recurring in nature.