Question

Find the maximum sum of the A.P \[40 + 38 + 36 + 34 + 32 + \_\_\_\_\_\_\_\]

Answer 1

Hint: A.P stands for arithmetic progression which is basically a sequence of numbers in which the difference of any two consecutive terms is same throughout. The given A.P is a decreasing collection of numbers where each successive term is smaller than the previous term. As we are asked to find the maximum of the sum we will find the sum up to when the term of the sequence is not negative. This can be achieved by using the formulas listed here.
Formulas used:
In an A.P the $ (n + 1) $ th term is represented as $ {a_n} $ , the common difference is denoted as $ d $ , the sum up to $ n $ terms is represented as $ {S_n} $ .
The formulas involved in an A.P are
1) $ {a_n} = {a_0} + (n - 1)d $
2) $ d = {a_{n + 1}} - {a_n} $
3) $ {S_k} = \sum\limits_{n = 0}^k {{a_n}} = \dfrac{k}{2} \times ({a_0} + {a_k}) $

Complete step-by-step answer:
The given A.P is \[40 + 38 + 36 + 34 + 32 + \_\_\_\_\_\_\_\]
Clearly we have $ {a_0} = 40,{a_1} = 38,{a_2} = 36,... $ and so on
The common difference is $ d = {a_{n + 1}} - {a_n} $
Let $ n = 1 $ so we have $ d = {a_2} - {a_1} = 38 - 40 = - 2 $
Since the common difference is a negative quantity and the value of $ {a_n} $ is decreasing as $ n $ increases so we can guess that the value of $ {a_n} = 0 $ for some value of $ n = k $ and it decreases to negative values too.
Now we have $ {a_k} = {a_0} + (k - 1)d $
We have the values $ {a_0} = 40,{a_k} = 0,d = - 2 $ so,
 $ 0 = 40 + (k - 1) \times ( - 2) $
 $ \Rightarrow - 40 = - 2 \times (k - 1) $
 $ \Rightarrow 20 = (k - 1) $
 $ \Rightarrow k = 20 + 1 = 21 $
Clearly, up to $ n = k = 21 $ we have all positive values of $ {a_n} $ and after that we have all negative values.
So the maximum sum for the A.P is $ {S_k} = \sum\limits_{n = 0}^k {{a_n}} $
By the formula for sum of an A.P we have,
 $ {S_{21}} = \sum\limits_{n = 0}^{21} {{a_n}} = \dfrac{{21}}{2} \times ({a_0} + {a_{21}}) $
\[ \Rightarrow {S_{21}} = \dfrac{{21}}{2} \times (40 + 0) = \dfrac{{21}}{2} \times 40\]
\[ \Rightarrow {S_{21}} = 21 \times 20 = 420\]
Therefore, the maximum sum of the given A.P is $ 420 $ .
So, the correct answer is “ $ 420 $”.

Note: The sum is calculated up to twenty terms to give the maximum value because after these many terms the A.P has only negative values, and if we add up the negative values to positive ones we will get a value less than the obtained one.