Answer :
- To show that the 18th term of an Arithmetic Progression (AP) is zero, let's start by using the general form of the nth term of an AP. The nth term ([tex]T_n[/tex]) of an AP is given by:
[tex]T_n = a + (n-1)d[/tex]
where [tex]a[/tex] is the first term and [tex]d[/tex] is the common difference.
For the 7th term, [tex]T_7[/tex], we have:
[tex]T_7 = a + 6d[/tex]
For the 11th term, [tex]T_{11}[/tex], we have:
[tex]T_{11} = a + 10d[/tex]
According to the problem, 7 times the 7th term is equal to 11 times the 11th term:
[tex]7(a + 6d) = 11(a + 10d)[/tex]
Expanding both sides:
[tex]7a + 42d = 11a + 110d[/tex]
Rearranging gives:
[tex]7a - 11a + 42d - 110d = 0[/tex]
[tex]-4a - 68d = 0[/tex]
Dividing the entire equation by -4:
[tex]a + 17d = 0[/tex]
Therefore, the 18th term [tex]T_{18}[/tex] is:
[tex]T_{18} = a + 17d[/tex]
From the previous result, since [tex]a + 17d = 0[/tex], the 18th term is:
[tex]T_{18} = 0[/tex]
Thus, the 18th term of the AP is zero.
- To find how many 3-digit numbers are divisible by 7, let's determine the smallest and largest 3-digit numbers that are divisible by 7.
The smallest 3-digit number is 100. To find the smallest 3-digit number divisible by 7, divide 100 by 7 and round up:
[tex]\frac{100}{7} \approx 14.2857[/tex]
Rounding up gives 15. So the smallest 3-digit number divisible by 7 is:
[tex]7 \times 15 = 105[/tex]
The largest 3-digit number is 999. To find the largest 3-digit number divisible by 7, divide 999 by 7 and round down:
[tex]\frac{999}{7} \approx 142.7143[/tex]
Rounding down gives 142. So the largest 3-digit number divisible by 7 is:
[tex]7 \times 142 = 994[/tex]
Now, using the arithmetic sequence formula to find how many such numbers exist:
This is an arithmetic sequence where the first term [tex]a_1[/tex] is 105, the last term [tex]l[/tex] is 994, and the common difference [tex]d[/tex] is 7.
The number of terms [tex]n[/tex] in an arithmetic sequence is given by:
[tex]n = \frac{l - a_1}{d} + 1[/tex]
Plugging in the values:
[tex]n = \frac{994 - 105}{7} + 1[/tex]
[tex]n = \frac{889}{7} + 1[/tex]
[tex]n = 127 + 1[/tex]
[tex]n = 128[/tex]
Therefore, there are 128 three-digit numbers that are divisible by 7.
