Question

The largest term common to sequence $1,11,21,31,....$ to 100 terms and $31,36,41,46,.....$ to 100 terms is
(A). 381
(B). 471
(C). 281
(D). None of these

Answer 1

Hint: You can observe that both the sequences are an arithmetic progression. So, look at the definition of arithmetic and find ${n^{th}}$ term formula. Find the relation between both sequences. If there is an arithmetic progression of first term a and common difference d, then t is written as $t = a + \left( {n - 1} \right)d$.

Complete step-by-step solution -
A sequence is of numbers such that the difference of any two consecutive numbers is a constant is called Arithmetic progression. For example, the sequence 1,2,3,4,……. Is an arithmetic progression with common difference $ = 2 - 1 = 1$. Now, we need to find ${n^{th}}$ term of such sequence. Let us have a sequence with first term a and common difference d.
We know the difference between successive terms is d.
So, $\text{second term} = \text{first term} + d = a + d$
$\text{third term} = \text{second term} + d = a + 2d$
And so on calculating, we get:
${n^{th}} = {\left( {n - 1} \right)^{th}}\,term\, + d = a + \left( {n - 2} \right)d + d = a + \left( {n - 1} \right)d$
Given sequence in the question is written in the form of:
$1,11,21,31,.......100\,\,terms$
Given the second sequence in the question is written in the form of:
$31,36,41,46,.......100\,\,terms$
Finding last term of both sequences, we should do the part:
For first sequence, $a = 1\,\,,\,\,d = 10,\,\,n = 100$
${n^{th}}$ term of A.P. with a first term “a”, common difference “d” is
$t = a + \left( {n - 1} \right)d$
So, here we need ${100^{th}}$ term of first sequence, which is given by : let it be denoted by t.
$t = a + \left( {n - 1} \right)d = 1 + \left( {100 - 1} \right)10$
By simplifying the above equation, we can say that the value of t:
$t = 1 + 990 = 991$
So, here we need ${100^{th}}$ term of second sequence, which is given by : let it be denoted by p.
$p = a + \left( {n - 1} \right)d$ Here $a = 31,\,\,n = 100,\,\,d = 5$
$p = 31 + 99 \times 5$
By simplifying we can write the value of p in the form of:
$p = 526$
As we can observe the ${100^{th}}$ term relation, we can say:
$t > p$.
The largest number which has the possibility to be in common:
$p = 526$
As you can observe the terms of the first sequence are odd. 526 cannot be present in the first sequence.
By observation, we can say every odd term i.e, alternate terms in second A.P. is present in first A.P.
So, if the last term of the second progression is not present in the first progression then we can say the last but one term of the second progression is definitely present in the first A.P. and it will be the largest possible.
(Last but on term of second) + (common difference) = last term
By substituting all and assuming required number as x, we get:
$x + 5 = 526$
By substituting 5 on both sides, we get the value of x as:
$x = 526 - 5 = 521$
This option is not there.
Therefore option (d) is the correct answer.

Note: Be careful while substituting ${n^{th}}$term formula, don’t mix up the terms of 2 progressions. The idea of saying every odd term of ${2^{nd}}$ is present in ${1^{st}}$ sequence is very crucial and base to the whole solution. Even though ‘t’ is large we took ‘p’ because the term must be present in both sequences. Observe the statement carefully.