Check whether 301 is a term of the list of numbers 5, 11, 17, 23, ...

Q: Check whether 301 is a term of the list of numbers 5, 11, 17, 23, ...

Given list of numbers: 5, 11, 17, 23, ... \nStep 1: Check if the sequence forms an AP. $11 - 5 = 6$ $17 - 11 = 6$ $23 - 17 = 6$\nSince the difference between consecutive terms is constant ($d = 6$), the given list is an AP with first term $a = 5$ and common difference $d = 6$. \nStep 2: Assume that 301 is the $n$-th term of this AP. Therefore, $an = 301$. \nStep 3: Use the $n$-th term formula. $an = a + (n - 1)d$ $301 = 5 + (n - 1) \times 6$ \nStep 4: Solve for $n$. $301 - 5 = 6(n - 1)$ $296 = 6(n - 1)$ $n - 1 = \frac{296}{6}$ $n - 1 = \frac{148}{3}$ $n = \frac{148}{3} + 1 = \frac{151}{3}$ \nStep 5: Analyze the value of $n$.\nSince $n$ comes out to be a fraction ($\frac{151}{3}$), it contradicts the condition that the number of terms ($n$) in a sequence must always be a positive integer. \nConclusion: Therefore, 301 is not a term of the given list of numbers.

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Given list of numbers: 5, 11, 17, 23, ... 
Step 1: Check if the sequence forms an AP. $11 - 5 = 6$ $17 - 11 = 6$ $23 - 17 = 6$
Since the difference between consecutive terms is constant ($d = 6$), the given list is an AP with first term $a = 5$ and common difference $d = 6$. 
Step 2: Assume that 301 is the $n$-th term of this AP. Therefore, $an = 301$. 
Step 3: Use the $n$-th term formula. $an = a + (n - 1)d$ $301 = 5 + (n - 1) 	imes 6$ 
Step 4: Solve for $n$. $301 - 5 = 6(n - 1)$ $296 = 6(n - 1)$ $n - 1 = \frac{296}{6}$ $n - 1 = \frac{148}{3}$ $n = \frac{148}{3} + 1 = \frac{151}{3}$ 
Step 5: Analyze the value of $n$.
Since $n$ comes out to be a fraction ($\frac{151}{3}$), it contradicts the condition that the number of terms ($n$) in a sequence must always be a positive integer. 
Conclusion: Therefore, 301 is not a term of the given list of numbers.

CBSE · Class 10 · Mathematics · Arithmetic ProgressionsCheck whether 301 is a term of the list of numbers 5, 11, 17, 23, ...