386 views

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. $16$

edited | 386 views
0
If you're uploading jpeg or any picture then make sure the alignment is right

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

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

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)
edited
0
thanks

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

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)
edited
0
thanks
0
"
∴ 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..
by Active (1.5k points)
+1

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

0
top secret