Question

Assertion :Between two numbers whose sum is $2\frac{1}{6},$ an even number of arithmetic means are inserted. If the sum of these means exceeds their number by unity, then the number of means are $12$ Reason: If $a$ and $b$ are two given numbers and $A_1,A_2,...,A_n$ are $n$ arithmetic means between them, then $A_1+A_2+...+A_n=n\left(\frac{a+b}{2}\right)$

• 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. 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

If $a$ and $b$ are two numbers , and $A_1,A_2,....A_n$ $n$ means are inserted between them then, total sum of numbers $=\frac{n+2}{2}(a+b)$

Sum of the arithmatic means inserted will be $=\text{sum of the numbers - (a+b)}$

$A_1+A_2+A_3+....+A_n=\frac{n+2}{2}(a+b)-(a+b)$

$=(a+b)\times \left(\frac{n}{2}\right)$

$\text{Sum}=n\left(\frac{a+b}{2}\right)$ .......(1)

$a+b=2\frac{1}{6}=\frac{13}{6}$

$\text{Sum}=n+1$

From (1), we have

$n\left(\frac{a+b}{2}\right)=n+1$

$\Rightarrow \frac{13n}{12}=n+1$

$\Rightarrow 13n=12n+12$

$\Rightarrow n=12$