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Question: If \(p^{th}\) terms of a list of number 21, 42, 63... is 420 then the value of \(p\) is \[\] A.20 ...

If pthp^{th} terms of a list of number 21, 42, 63... is 420 then the value of pp is A.20 A.20
B.21 C.23 C.23
D.19 $$$$

Explanation

Solution

find the first term and the common difference in the given sequence of numbers. Use the formula of nthn^{th} term of an arithmetic sequence to put the obtained and given values. Solve the resulting equation to get the value of pp $$$$

Complete step by step answer:
A sequence is defined as the enumerated collection of numbers where repetitions are allowed and order of the numbers matters. It can also be expressed as a one-one map from the natural numbers set to real numbers. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
(xn)=x1,x2,x3,...\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...
The given sequence in the question is 21, 42, 63,.. We see that the given sequence is infinite and contains the first three numbers 21, 42, 63. As given in the question that three numbers follow some particular rule and also the same rule applies for 420 as 420 is also present in the sequence. Letustrytofindarelationamongthefirstthreeterms.Weknowtwospecialsequencearithmeticsequences(AP)and(geometricsequence)GP.APsequencehasthesamedifferencebetweentwoconsecutiveterms.Letuscheckinthesequence. Let us try to find a relation among the first three terms. We know two special sequence arithmetic sequences (AP) and (geometric sequence ) GP. AP sequence has the same difference between two consecutive terms. Let us check in the sequence.
We see 42-21=63-42=21. We know that the nthn^{th} term of an AP with first term aa and common difference dd is given by
xn=a+(n−1)d{{x}_{n}}=a+\left( n-1 \right)d
We see in this sequence the first term is a=21a=21, the common difference is d=42−21=63−42=21d=42-21=63-42=21 , n=pn=p and nthn^{th} term is 420. Let us put the obtained values in the formula and get ,

& {{x}_{n}}=a+\left( n-1 \right)d \\\ & \Rightarrow 420=21+\left( p-1 \right)21 \\\ & \Rightarrow \dfrac{420-21}{21}=\left( p-1 \right) \\\ & \Rightarrow p-1=19 \\\ & \Rightarrow p=20 \\\ \end{aligned}$$ **So, the correct answer is “Option A”.** **Note:** We need to be careful of the confusion between the $n^{th}$ of an AP and GP. You can also solve the problem by observing that all numbers in the list are integral multiples of 21 . So 420 will be $\dfrac{420}{21}=20^{th}$ term of the sequence.