Question

The largest term common to the sequence 1, 11, 21, 31 to 100 terms and 31, 36, 41, 46 to 100 terms is,
a) 381
b) 471
c) 281
d) 521

Answer 1

Hint: Find the \[{{100}^{th}}\] term of both the sequences after checking if it is in AP or not. The largest number of sequence 2 will be a term in sequence 1. Find the largest term corresponding to that value.

Complete step-by-step answer:
Here we need to find the largest term which is common to both the sequence.
Sequence 1 \[\Rightarrow \] 1, 11, 21, 31,…….. to 100 terms.
Sequence 2 \[\Rightarrow \] 31, 36, 41,….... to 100 terms.
Let us consider sequence 1, it is in Arithmetic Progression (AP) 1, 11, 21, 31,…… to 100 tems.
An AP is a sequence of numbers in which each term is derived from the preceding term by addition or subtraction, having the common difference ‘d’.
Here common difference, ‘d’ = \[{{2}^{nd}}\] term - \[{{1}^{st}}\] term = 11 – 1 =10.
Similarly, d = \[{{3}^{rd}}\] term - \[{{2}^{nd}}\] term = 21 – 11 = 10.
\[\therefore \] d = 10, first term, a = 1.
To find the \[{{n}^{th}}\] term of a series, we use the formula,
\[{{T}_{n}}=a+\left( n-1 \right)d\]
So to get \[{{100}^{th}}\] term of the given AP series,
\[\begin{align}
  & {{T}_{100}}=1+\left( 100-1 \right)\times 10 \\
 & {{T}_{100}}=1+99\times 10=1+990=991 \\
\end{align}\]
\[\therefore \] \[{{100}^{th}}\] term of Sequence 1 = 991 = \[{{T}_{100}}\].
Similarly if we are considering the \[{{2}^{nd}}\] sequence,
31, 36, 41, 46,….. to 100 terms.
This sequence is also in AP, with first term, a = 31.
Common difference, d = \[{{2}^{nd}}\] term - \[{{1}^{st}}\] term = 36 – 31 =5.
Similarly, d = \[{{3}^{rd}}\] term - \[{{2}^{nd}}\] term = 41 – 36 = 5.
Hence, d = 5, a = 31.
Let us find \[{{100}^{th}}\] term of sequence 2.
\[\begin{align}
  & {{T}_{100}}=a+\left( n-1 \right)d \\
 & {{T}_{100}}=31+\left( 100-1 \right)\times 5 \\
 & {{T}_{100}}=31+\left( 99\times 5 \right)=526 \\
\end{align}\]
Let 526 be the largest term in both the series. Now we need to find the common term as in the sequence 1.
\[\begin{align}
  & 526=31+\left( n-1 \right)\times 10 \\
 & \Rightarrow 526-31=10n-10 \\
 & \therefore 10n=495+10 \\
 & 10n=505 \\
 & \therefore n=50.5 \\
\end{align}\]
The term cannot be an integer, so we can round it to value 50.
\[\therefore \] n = 50.
Now let us find the value of \[{{T}_{50}}\],
\[\begin{align}
  & {{T}_{50}}=31+\left( 50-1 \right)\times 10 \\
 & {{T}_{50}}=31+49\times 10 \\
 & {{T}_{50}}=31+490=521 \\
\end{align}\]
Hence we have found the largest common term in both series as 521.
\[\therefore \] Option (d) is the correct answer.

Note: In a question like this first check if it's AP or GP. Then use the formula to bring out a relation between both the sequences and then try to find the common terms between them. We found \[{{T}_{{{100}^{th}}}}\] term because knowing \[{{1}^{st}}\] and last term will help us to find the required term.