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The number of common terms in  the two sequences $\{ 3,7,11, \ldots , 407\}$ and $\{2,9,16,\ldots ,709\}$ is

  1. $13$
  2. $14$
  3. $15$
  4. $16$
in Numerical Ability by Active (1.5k points)
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3 Answers

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$S_1 = \begin{Bmatrix} 3,7,11,15,19,23,\ldots,407 \end{Bmatrix}$
$S_2 = \begin{Bmatrix} 2,9,16,23,\ldots,709 \end{Bmatrix}$

$\text{common diff} (d_1) \text{ for }  S_1 = 4$
$\text{common diff} (d_2) \text{ for }  S_2 = 7$

Now, we try to make a new sequence which contains the common elements of the given two arithmetic progression $(AP)$ sequences.

$\text{first common term}(a) = 23$, this will be the first term in our new sequence which will also be an AP sequence and its common difference,

$d = LCM(d_1 , d_2) = LCM(4,7) = 28.$

$n^{th}$ term of our new sequence is given by,

$T_n = a + (n-1)*d$
$\quad = 23 + (n-1)*28$
$\quad = 23 + 28n - 28$
$\quad = 28n - 5$

For the last term of our sequence we get,

$28n - 5 = \min(407,709) = 407$

$\implies 28n = 407 + 5 = 412$

$n = \dfrac{412}{28}= 14.741$

$n$ must be integer and since we are finding number of present terms we must take $floor$ meaning the last term can be $407$ or some value lower than it. So, $n = 14.$

There are $14$ common terms.

Correct Answer: $B$
by Boss (10.8k points)
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+3 votes

$\color{violet}{\text{1st method :}}$

Let, the number of terms present in the first sequence be $m$

We know, $a+(m-1)d = x_m$

where, a = first term

d = common difference

m = no. of terms

$x_m=$ mth term

∴ $3+(m-1) \times 4 = 407$

Or, $3+4m-4 = 407$

Or, $4m = 407+1 = 408$

Or, $m = 102$

Let, the number of terms present in the second sequence be $n$

∴ $a+(n-1)d = x_n$

where, a = first term

d = common difference

n = no. of terms

$x_n=$ nth term

∴ $2+(n-1) \times 7 = 709$

Or, $2+7n-7 = 709$

Or, $7n = 709+5 = 714$

Or, $n = 102$

First common term between two sequence is $23$

next common term will be $51$

& next common term will be $79$

.......................

..........

∴ Common difference will be $(51-23) = (79-51)  28$

& Last common term will be less than $407$ or equal to $407$

∴ $a+(p-1)d = x_p$

where, a = first term

d = common difference

p = no. of terms

$x_p=$ pth term

Or, $23+ (p-1) \times 28 \leq 407$

Or, $23+28p -28 \leq  407$

Or, $28p \leq 407+5 = 412$

Or, $p \leq \dfrac{412}{28} $

Or, $p \leq 14.714$

We'll take only 14 .

So, The number of terms in two sequences $\{ 3,7,11,.....,407\}$ and $\{ 2,9,16, .... , 709\}$ is $14$

$\color{violet}{\text{2nd method :}}\color{maroon}{\text{ (easy one)}}$

The common difference of the first series is 4

The common difference of the second series is 7

lcm(common difference of the first series, common difference of the second series) will be the common difference of the common terms series.

∴ $lcm(4,7) = 28$

 

First common term between two sequence is 23

next common term will be 51

& next common term will be 79

∴ Common difference will be $(51-23) = (79-51)  28$

& Last common term will be less than $407$ or equal to $407$

∴ $a+(p-1)d = x_p$

where, a = first term

d = common difference

p = no. of terms

$x_p=$ pth term

Or, $23+ (p-1) \times 28 \leq 407$

Or, $23+28p -28 \leq  407$

Or, $28p \leq 407+5 = 412$

Or, $p \leq \dfrac{412}{28} $

Or, $p \leq 14.714$

We'll take only 14 .

So, The number of terms in two sequences $\{ 3,7,11,.....,407\}$ and $\{ 2,9,16, .... , 709\}$ is $14$

by Boss (17.8k points)
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"
∴ lcm(common difference of the first series, common difference of the second series) will be the common difference of the common terms series".

Mam can you explain this part..
0 votes
ANSWER is (B)
by Active (1.5k points)
+1

jjayantamahata : Why are you just writing the option as an answer instead of providing the full solution?

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