Question

Assertion: The sum of the series with the ${{n}^{th}}$ term, ${{t}_{n}}=\left( 9-5n \right)$ is $-465$, when no. of terms $n=15$.
Reason: Given series in A.P. and sum of $n$ terms of an A.P. is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.
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: To solve this question, we have some given values such that the ${{n}^{th}}$ term and the sum of $n$ terms and the number of terms. Now, we will put the value of $n$ to get the terms of the series and find the common difference of two consecutive numbers and use the formula of sum of $n$ numbers to verify the reason.

Complete step by step solution:
Given that:
$\Rightarrow $${{n}^{th}}$ term ${{t}_{n}}=\left( 9-5n \right)$
$\Rightarrow $Sum of $n$ terms $=-465$
$\Rightarrow $Number of terms, $n=15$
For first term, $n=1$
$\begin{align}
  & \Rightarrow {{t}_{1}}=\left( 9-5\times 1 \right) \\
 & \Rightarrow {{t}_{1}}=\left( 9-5 \right) \\
 & \Rightarrow {{t}_{1}}=4 \\
\end{align}$
For second term, $n=2$
$\begin{align}
  & \Rightarrow {{t}_{2}}=\left( 9-5\times 2 \right) \\
 & \Rightarrow {{t}_{2}}=\left( 9-10 \right) \\
 & \Rightarrow {{t}_{2}}=-1 \\
\end{align}$
For third term, $n=3$
$\begin{align}
  & \Rightarrow {{t}_{3}}=\left( 9-5\times 3 \right) \\
 & \Rightarrow {{t}_{3}}=\left( 9-15 \right) \\
 & \Rightarrow {{t}_{3}}=-6 \\
\end{align}$
For fourth term, $n=4$
$\begin{align}
  & \Rightarrow {{t}_{4}}=\left( 9-5\times 4 \right) \\
 & \Rightarrow {{t}_{4}}=\left( 9-20 \right) \\
 & \Rightarrow {{t}_{4}}=-11 \\
\end{align}$
For fifth term, $n=5$
$\begin{align}
  & \Rightarrow {{t}_{5}}=\left( 9-5\times 5 \right) \\
 & \Rightarrow {{t}_{5}}=\left( 9-25 \right) \\
 & \Rightarrow {{t}_{5}}=-16 \\
\end{align}$
For sixth term, $n=6$
$\begin{align}
  & \Rightarrow {{t}_{6}}=\left( 9-5\times 6 \right) \\
 & \Rightarrow {{t}_{6}}=\left( 9-30 \right) \\
 & \Rightarrow {{t}_{6}}=-21 \\
\end{align}$
For seventh term, $n=7$
$\begin{align}
  & \Rightarrow {{t}_{7}}=\left( 9-5\times 7 \right) \\
 & \Rightarrow {{t}_{7}}=\left( 9-35 \right) \\
 & \Rightarrow {{t}_{7}}=-26 \\
\end{align}$
For eighth term, $n=8$
$\begin{align}
  & \Rightarrow {{t}_{8}}=\left( 9-5\times 8 \right) \\
 & \Rightarrow {{t}_{8}}=\left( 9-40 \right) \\
 & \Rightarrow {{t}_{8}}=-31 \\
\end{align}$
For ninth term, $n=9$
$\begin{align}
  & \Rightarrow {{t}_{9}}=\left( 9-5\times 9 \right) \\
 & \Rightarrow {{t}_{9}}=\left( 9-45 \right) \\
 & \Rightarrow {{t}_{9}}=-36 \\
\end{align}$
For tenth term, $n=10$
$\begin{align}
  & \Rightarrow {{t}_{10}}=\left( 9-5\times 10 \right) \\
 & \Rightarrow {{t}_{10}}=\left( 9-50 \right) \\
 & \Rightarrow {{t}_{10}}=-41 \\
\end{align}$
For eleventh term, $n=11$
$\begin{align}
  & \Rightarrow {{t}_{11}}=\left( 9-5\times 11 \right) \\
 & \Rightarrow {{t}_{11}}=\left( 9-51 \right) \\
 & \Rightarrow {{t}_{11}}=-46 \\
\end{align}$
For twelfth term, $n=12$
$\begin{align}
  & \Rightarrow {{t}_{12}}=\left( 9-5\times 12 \right) \\
 & \Rightarrow {{t}_{12}}=\left( 9-60 \right) \\
 & \Rightarrow {{t}_{12}}=-51 \\
\end{align}$
For thirteenth term, $n=13$
$\begin{align}
  & \Rightarrow {{t}_{13}}=\left( 9-5\times 13 \right) \\
 & \Rightarrow {{t}_{13}}=\left( 9-65 \right) \\
 & \Rightarrow {{t}_{13}}=-56 \\
\end{align}$
For fourteenth term, $n=14$
$\begin{align}
  & \Rightarrow {{t}_{14}}=\left( 9-5\times 14 \right) \\
 & \Rightarrow {{t}_{14}}=\left( 9-70 \right) \\
 & \Rightarrow {{t}_{14}}=-61 \\
\end{align}$
For fifteenth term, $n=15$
$\begin{align}
  & \Rightarrow {{t}_{15}}=\left( 9-5\times 15 \right) \\
 & \Rightarrow {{t}_{15}}=\left( 9-75 \right) \\
 & \Rightarrow {{t}_{15}}=-66 \\
\end{align}$
Here, we got the $15$ terms of the series. Now, we will check the common difference between two consecutive terms as:
\[\begin{align}
  & \Rightarrow {{t}_{2}}-{{t}_{1}}={{t}_{3}}-{{t}_{2}}={{t}_{4}}-{{t}_{3}}=...={{t}_{15}}-{{t}_{14}} \\
 & \Rightarrow -1-4=-6-\left( -1 \right)=-11-\left( -6 \right)=...=-66-\left( -61 \right) \\
 & \Rightarrow -5=-5=-5=...=-5 \\
\end{align}\]
As it can be observed that common differences between two consecutive terms are constant. So, the given series is in A.P.
Since, the given series is in A.P. So, for calculating the sum of $n$ terms, we can use the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where ${{S}_{n}}$ is sum of $n$ terms, $n$ is number of terms, $a$ is the first term of the series and $d$ is the common difference between two consecutive terms.

Note: We will check the solution by using formula of sum of A.P. as:
Since, we have the sum of $15$ terms is $-465$. And we got $a=4$ and $d=-5$ from the solution. So, we will substitute it in the formula as:
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\
 & \Rightarrow -465=\dfrac{15}{2}\left[ 2\times 4+\left( 15-1 \right)\left( -5 \right) \right] \\
\end{align}$
Simplify the equation as:
$\begin{align}
  & \Rightarrow -465=\dfrac{15}{2}\left[ 2\times 4+\left( 15-1 \right)\left( -5 \right) \right] \\
 & \Rightarrow -465=\dfrac{15}{2}\left[ 8+14\left( -5 \right) \right] \\
 & \Rightarrow -465=\dfrac{15}{2}\left[ 8-70 \right] \\
 & \Rightarrow -465=\dfrac{15}{2}\left( -62 \right) \\
 & \Rightarrow -465=15\times \left( -31 \right) \\
 & \Rightarrow -465=-465 \\
\end{align}$
As we can see $L.H.S.=R.H.S.$
Hence, the solution is correct.