Question

Assertion: $ n^{th} $ term $ ({T_n}) $ of the sequence $ (1,6,18,40,75,126,....) $ is an $ a{n^3} + b{n^2} + cn + d $ and $ 6a + 2b - d $ is $ 4 $ .
Reason: If the second successive difference (Differences of the differences) of a series are in A.P, then $ {T_n} $ is a cubic polynomial is n.
A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
C. Assertion is correct but Reason is incorrect.
D. Both Assertion and Reason are incorrect.

Answer 1

Hint: There are two statements given assertion and reason, and we need to check if both statements are correct or wrong.
For that, we need to know about the Arithmetic progression. An arithmetic progression can be given by $ a,(a + d),(a + 2d),(a + 3d),... $ where $ a $ is the first term and $ d $ is a common difference.
Formula used: Consider for solving these questions $ {a_n} = a + (n - 1)d $
Where $ d $ is the common difference, $ a $ is the first term, since we know that the difference between consecutive terms is constant in any A.P.
For the cubic polynomial $ P(n) = a{n^3} + b{n^2} + cn + d $ .

Complete step by step answer:
Let the sequence is given as $ (1,6,18,40,75,126,....) $ .
Now we will find the first difference of the given series by taking the terms of the common difference next to it like $ 6 - 1 = 5 $ and $ 18 - 6 = 12 $ similarly.
Hence the first set of the difference series is $ 5,12,22,35 $ .
We can also find the second different set of the series using the same method as above but using series one; thus, we get $ 7,10,13 $ where $ 12 - 5 = 7,22 - 12 = 10,35 - 22 = 13 $ .
Hence, we get the Arithmetic progression on the second difference series; where a is seven and d is three.
Also, the general terms of the given series are cubic because it contains three values only.
For the cubic polynomial, we know that $ P(n) = a{n^3} + b{n^2} + cn + d $ .
Now we are going to replace the cubic polynomial with the values of one, two, three, and four.
 $ P(1) = a + b + c + d = 1 $ (right side one is the first term in the given polynomial) $ ---(1) $
 $ P(2) = 8a + 4b + 2c + d = 6 $ $ ---(2) $
 $ P(3) = 27a + 9b + 3c + d = 18 $ $ --(3) $
 $ P(4) = 64a + 16b + 4c + d = 40 $ $ --(4) $
Thus, we have the four equations of the cubic polynomial.
First equation $ (2) $ - $ (1) $ we get $ 7a + 3b + c = 5 $ $ --(5) $
Second equation $ (3) $ - $ (2) $ we get $ 26a + 8b + 2c = 17 $ $ ---(6) $
Finally, equation $ (4) $ - $ (3) $ we get $ 63a + 15b + 3c = 39 $ $ --(7) $
Hence solving these three: five, six, and seventh equation we get; $ a = \dfrac{1}{2},b = \dfrac{1}{2},c = 0,d = 0 $
Hence from the assertion $ 6a + 2b - d = 6(\dfrac{1}{2}) + 2(\dfrac{1}{2}) - 0 $ . Thus, we get $ 6a + 2b - d = 4 $ .

Hence both assertion and reason are correct.

Note: Also, the reason is the correct explanation for the given assertion.
Since a cubic polynomial is of power at most three only thus $ P(n) = a{n^3} + b{n^2} + cn + d $ .
There is also another progression, which is geometric progression and terms are $ \dfrac{a}{r},a,ar,.... $ where a is the first term and r is the ratio.