Question

Assertion :In an $A.P.$ of odd number of terms, let $S_1$ & $S_2$ are such that $S_1=t_1+t_2+t_3+...+t_n$ and $S_2=t_1+t_3+t_5+...+t_n$ then $\frac{s_1}{s_2}=\frac{n}{n+1}$ Reason: If $1,2,3,...,n$ be the numbers where $n$ is odd then $1,3,5,7...n$ will be $\frac{n+1}{2}$ odd numbers & $2,4,6....(n-1)$ will be $\frac{n^2-1}{2}$ even numbers.

• 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 D Assertion is incorrect but Reason is correct

If there are 9 terms then $t_2,t_4,t_6,t_8$ will be $\frac{9-1}{2}=4$

in numbers & $t_1,t_3,t_5,t_7,t_9$ will be $\frac{9+1}{2}=5$ in numbers.

Hence $S_1$ be an $A.P.$ of $n$ terms with common difference $2d$

but $S_2$ be an $A.P.$ of $\frac{n+1}{2}$terms with common difference $2d$.

Now $S_1=\frac{n}{2}[2a+(n-1)d]$

$S_2=\frac{1}{2}\left(\frac{n+1}{2}\right)[2a+(\frac{n+1}{2}-1)2d]=\frac{n+1}{4}[2a+(n-1)d]$

$\therefore \frac{S_1}{S_2}=\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n+1}{4}[2a+(n-1)d]}=\frac{2n}{n+1}$

$\therefore$ Assertion (A) is false but Reason (R) is true