Question

The consecutive odd integers whose sum is ${{45}^{2}}-{{21}^{2}}$ are
(a) 43, 45,…………, 75
(b) 43, 45, ............, 79
(c) 43, 45, ……………, 85
(d) 43, 45, ……………., 89

Answer 1

We solve this problem by using the standard formula of sum of first $'n'$ odd natural numbers is given as ${{n}^{2}}$that is the general form of odd number is $2p-1$ then
$\sum\limits_{p=1}^{n}{\left( 2p-1 \right)}={{n}^{2}}$
Here the last odd number in the above sum will be $2n-1$
We use the above result to find the starting term and ending term whose sum will be ${{45}^{2}}-{{21}^{2}}$ to find the consecutive odd numbers.

Complete step-by-step solution
We are given that the sum of some consecutive odd numbers is ${{45}^{2}}-{{21}^{2}}$
We know that the standard formula of sum of first $'n'$ odd natural numbers is given as ${{n}^{2}}$that is the general form of odd number is $2p-1$ then
$\sum\limits_{p=1}^{n}{\left( 2p-1 \right)}={{n}^{2}}$
Here the last odd number in the above sum will be $2n-1$
Let us assume that the sum of $'{{n}_{1}}'$ odd natural numbers as ${{45}^{2}}$ then we get
$\Rightarrow \sum\limits_{p=1}^{{{n}_{1}}}{\left( 2p-1 \right)}={{45}^{2}}$
Let us assume that the last odd number in this sequence as $'a'$
Here we can see that the last odd number in the above equation is given as

& \Rightarrow a=2\left( 45 \right)-1 \\\ & \Rightarrow a=89 \\\ \end{aligned}$$ Let us assume that the sum of $$'{{n}_{2}}'$$ odd natural numbers as $${{21}^{2}}$$ then we get $$\Rightarrow \sum\limits_{p=1}^{{{n}_{2}}}{\left( 2p-1 \right)}={{21}^{2}}$$ Let us assume that the last odd number in this sequence as $$'b'$$ Here we can see that the last odd number in the above equation is given as $$\begin{aligned} & \Rightarrow b=2\left( 21 \right)-1 \\\ & \Rightarrow b=41 \\\ \end{aligned}$$ Now, we can see that the sum of odd numbers from 1 to 89 is $${{45}^{2}}$$ and sum of odd numbers from 1 to $$41$$ is $${{21}^{2}}$$ Therefore, we can conclude that the odd numbers having sum $${{45}^{2}}-{{21}^{2}}$$ is given by subtracting sum of odd numbers from 1 to 41 from sum of odd numbers from 1 to 89 $$\Rightarrow {{45}^{2}}-{{21}^{2}}=\left( \text{sum of odd numbers from 1 to 89} \right)-\left( \text{sum of odd numbers from 1 to 41} \right)$$ By using the numbers to above equation we get $$\begin{aligned} & \Rightarrow {{45}^{2}}-{{21}^{2}}=\left( 1+3+5+......+41+43+45+.....+89 \right)-\left( 1+3+5+....+41 \right) \\\ & \Rightarrow {{45}^{2}}-{{21}^{2}}=43+45+.......+89 \\\ \end{aligned}$$ Therefore the consecutive odd numbers whose sum is $${{45}^{2}}-{{21}^{2}}$$ are 43, 45, ……, 89 **So, option (d) is the correct answer.** **Note:** We can solve this problem by checking the options. We are given that the sum of odd numbers as $$\begin{aligned} & \Rightarrow S={{45}^{2}}-{{21}^{2}} \\\ & \Rightarrow S=2025-441 \\\ & \Rightarrow S=1584 \\\ \end{aligned}$$ Let us check the option (d) that is 43, 45, ….., 89 Let us assume that the sum these odd numbers as $$N$$ that is $$\Rightarrow N=43+45+.......+89$$ Here we can see that the above sequence is in A.P with first term as 43 and last term as 89 and number of terms as 24 We know that the formula for sum of $$'n'$$ terms of A.P having first term as $$'{{a}_{1}}'$$ and the last term as $$'{{a}_{n}}'$$ is given as $$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)$$ By using the above formula we get $$\begin{aligned} & \Rightarrow N=\dfrac{24}{2}\left( 43+89 \right) \\\ & \Rightarrow N=12\times 132 \\\ & \Rightarrow N=1584 \\\ \end{aligned}$$ Here we can see that the sum we found is equal to the sum given in the question. So, option (d) is the correct answer.