Question

Answered question n.o 69

Answer 1

Let me show you a 'logical' short method!

Consider the last terms in both the APs - 399 and 497.

399 < 497

From both APs, it is seen that '7' is common.

Now, take the common difference of both APs.

Common difference of first AP = 7 - 3 = 4.

Common difference of second AP = 7 - 2 = 5.

Then take the LCM of both the common differences.

4 × 5 = 20.

Here I got that, to the lowest common term, if 20 is added, then we get next common term.

Let's check.

⇒ Lowest common term 7.

⇒ 7 + 20 = 27 is also common.

⇒ So is 27 + 20 = 47.

⇒ So is 47 + 20 = 67.

⇒ So is 67 + 20 = 87.

(1)  Here the common terms are also in an AP.

7, 27, 47, 67, 87,...

(2)  And the no. formed by the digits in these numbers except 7 are consecutive even integers.

07, 27, 47, 67, 87, 107, 127,...

So,  as 399 < 497, the largest common term in both APs is less than 399.

According to (2), the largest common term in both APs is 387.

After 387, the common term would be 407 but it's not in first AP 3, 7, 11, 15, 19,...

Now, consider the AP 7, 27, 47,......, 387 thus formed.

We have to find the no. of terms in this AP, which is the answer.

$a=7 \\ \\ d=20 \\ \\ a_n=387 \\ \\ \\ \\ n=\displaystyle \frac{a_n-a}{d}+1 \\ \\ \\ n=\frac{387-7}{20}+1 \\ \\ \\ n=\frac{380}{20}+1 \\ \\ \\ n=19+1 \\ \\ \\ n=\bold{20}$

Thus there are 20 common terms.

So option (3) is the answer.

Another method...

First we have to find the algebraic form of both APs.

Consider first AP 3, 7, 11,...

$a=3 \\ \\ d=7-3=4 \\ \\ \\ a_n=dn+a-d \\ \\ a_n=4n+3-4 \\ \\ a_n=4n-1$

Consider second AP 2, 7, 12, 17,...

$a=2 \\ \\ d=7-2=5 \\ \\ \\ a_n=dn+a-d \\ \\ a_n=5n+2-5 \\ \\ a_n=5n-3$

To find the common terms, let the x'th term of 1st AP and y'th term of 2nd AP are same.

Thus,

$4x-1=5y-3 \\ \\ 5y-4x=3-1 \\ \\ 5y-4x=2$

This shows that the position numbers of each common term in both APs satisfy this equation.

$5y-2=4x$

Consider 5y. It always ends in either 5 or 0.

As 4x is obtained by subtracting 2 from 5y,  4x should end in either 3 or 8 respectively. But there's no multiple of 4 which ends in 3.

Thus there's only 1 possibility for 4x and 5y each: the ones digit of 4x and 5y should be 8 and 0 respectively.

So, a few possible numbers as values of x and y, the values of 4x and 5y and the value of 4x - 5y are tabled below.

$\begin{tabular}{|c|c|c|c|c|}\cline{1-5} x & y & 4x & 5y & 5y-4x \\ \cline{1-5}\cline{1-5}2&2&8&10&2\\ \cline{1-5}7&6&28&30&2\\ \cline{1-5}12&10&48&50&2\\ \cline{1-5}17&14&68&70&2\\ \cline{1-5}22&18&88&90&2\\ \cline{1-5}\end{tabular}$

The values of x and y each are also in APs.

Values of x = 2, 7, 12, 17,...

Values of y = 2, 6, 10, 14,...

Now we have to find the position number of last term in the AP 3, 7, 11,... i.e., 399, and then find the largest possible value for x.

Consider the AP 3, 7, 11,...

$a_n=4n-1 \\ \\ 399=4n-1 \\ \\ 4n=399+1 \\ \\ 4n=400 \\ \\ n=100$

Now consider the AP 2, 7, 12,... whose terms are possible values of x.

$a=2 \\ \\ d=7-2=5 \\ \\ a_n=dn+a-d \\ \\ a_n=5n+2-5 \\ \\ a_n=5n-3$

Assume 100 is a term of this AP.

$5n-3=100 \\ \\ 5n=103 \\ \\ n=20.6$

The whole number before n is 20. So we've to find the 20th term in this AP.

$5n-3 = 5 \times 20-3=100-3=97$

Thus 97 is the largest possible value for x.

Now let's find the 20th term of the AP 2, 6, 10, 14,... whose terms are possible values of y.

$a=2 \\ \\ d=6-2=4 \\ \\ a_n=dn+a-d \\ \\ a_n=4n+2-4 \\ \\ a_n=4n-2$

$a_{20}=4 \times 20-2\\ \\ a_{20}=80-2 \\ \\ a_{20}=78$

Thus 78 is the largest possible value for y.

No. of terms in both APs 2, 7, 12, 17,......, 97 and 2, 6, 10, 14, 18,......, 78 would be equal, and that's the answer.

$\\ \\ \\ \displaystyle \frac{97-2}{5}+1 \ = \ \frac{78-2}{4}+1 \\ \\ \\ \frac{95}{5}+1 \ = \ \frac{76}{4}+1 \\ \\ \\ 19+1 \ = \ 19+1 \\ \\ \\ \bold{20\ =\ 20}$

Thus 20 is the answer, i.e., option (3).