Question

Sum of certain consecutive odd positive integers is \[{57^2} - {13^2}\]. Find the first integer of that series.
A.29
B.27
C.25
D.31

Answer 1

Hint: : Here we have to find the value of the first integer of the series. We will first assume the odd consecutive numbers to be any variable. Then we will use the sum formula of an A.P to find the sum of the series as the given series is forming an A.P. We will then equate the sum with the given sum to get an equation. We will solve the equation to find the required values.

Formula Used:
We will use the formula of sum of an A.P , \[\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\] , where \[n\] is the number of terms in series, \[a\] is the first term, \[d\] is the common difference.

Complete step-by-step answer:
Let the odd integers be \[2k + 1\], \[2k + 3\], \[2k + 5\] ……
Let the number of odd positive integers be \[n\].
This series is forming an A.P.
The first term of the given series is \[2k + 1\] and the common difference is \[2k + 3 - \left( {2k + 1} \right) = 2\]
Let the sum of the given series is \[{S_n}\].
Applying the sum formula for the given series, \[\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], we get
\[ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2\left( {2k + 1} \right) + \left( {n - 1} \right)2} \right)\]
Simplifying the terms inside the bracket, we get
\[ \Rightarrow {S_n} = n\left( {2k + n} \right)\]
Using the distributive property of multiplication, we get
\[ \Rightarrow {S_n} = 2kn + {n^2}\]
Adding and the subtracting the term \[{k^2}\], we get
\[ \Rightarrow {S_n} = 2kn + {n^2} + {k^2} - {k^2}\]
Using algebraic identities \[{a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}\] in the above equation, we get
Therefore,
\[ \Rightarrow {S_n} = {\left( {n + k} \right)^2} - {k^2}\] …….. \[\left( 1 \right)\]
The sum of the series is \[{57^2} - {13^2}\]. Equating the value of sum in the equation \[\left( 1 \right)\] with the given sum, we get
\[ \Rightarrow {57^2} - {13^2} = {\left( {n + k} \right)^2} - {k^2}\]
Equating the terms, we get
\[57 = n + k\] and \[k = 13\]
Therefore, the value of \[n\] is \[57 - 13 = 44\]
We know that the first integer of the given series is \[2k + 1\]
We will substitute the value of \[k\] here.
\[ \Rightarrow 2k + 1 = 2 \times 13 + 1 = 27\]
Therefore, the first odd integer is 27.
Hence, the correct option is B.

Note: Here we have observed that the consecutive odd numbers is forming an arithmetic progression here. Similarly, series including the consecutive even numbers will also form an arithmetic progression. An arithmetic series is defined as a group of numbers whose difference of the terms from the preceding term is constant.
We have taken \[2k + 1\], \[2k + 3\], \[2k + 5\] …… as our series for odd integers because there is a difference of 2 between two consecutive odd numbers. If we take \[2k + 1\], \[2k + 2\], \[2k + 3\] ……, then the series will have both even and odd numbers and not particularly odd numbers.