Question

Assertion :$n^{th}$ term $\left(T_n\right)$ of the sequence (1, 6, 18, 40, 75, 126 ......) is $a{n}^3+b{n}^2+cn+d$. and 6a + 2b - d is = 4. Reason: If the second successive differences (Differences of the differences) of a series are in A.P., then $T_n$ is a cubic polynomial in 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. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$1,6,18,40,\mathrm{75...}$
First difference series$=5,12,22,35$
Second difference series$=7,10,13$
Hence the second difference is in A.P.
Therefore the general term of the series is cubic.
Now let the cubic be
$P\left(n\right)=a{x}^3+b{x}^2+cx+d$
Therefore we need 4 equations to find the values of a,b,c,d.
Hence
$P\left(1\right)=a+b+c+d=1$ ...(i)
$P\left(2\right)=8a+4b+2c+d=6$ ...(ii)
$P\left(3\right)=27a+9b+3c+d=18$...(iii)
$P\left(4\right)=64a+16b+4c+d=40$ ....(iv)
$ii-i$ implies
$7a+3b+c=5$ ...(v)
$iii-i$ implies
$26a+8b+2c=17$ ...(vi)
$Iv-i$ implies
$63a+15b+3c=39$ ...(vii)
Solving $v,vi,vii$ we get
$a=\frac{1}{2}$, $b=\frac{1}{2}$, $c=0$.
Substituting in i, we get
$d=0$.
Hence
$6a+2b-d$
$=6\left(\frac{1}{2}\right)+2\left(\frac{1}{2}\right)-0$
$=4$.
Hence both assertion and reasons are true, and reason is the correct explanation of assertion.