# a) If pth term of an A.P is q and qth term of the A.P. is p, find its rth term. b) Find the sum of all the integers between 100 and 1000 that are divisible by 9.

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a) If $$p$$th term of an A.P is $$q$$ and $$q$$th term of the A.P. is $$p$$, find its $$r$$th term:

In an arithmetic progression (A.P.), the $$n$$th term ($$a_n$$) is given by:

$a_n = a + (n-1)d$

where $$a$$ is the first term and $$d$$ is the common difference.

Given that the $$p$$th term is $$q$$, we can write:

$a_p = a + (p-1)d = q$

Similarly, for the $$q$$th term:

$a_q = a + (q-1)d = p$

Now, let's solve these two equations simultaneously. Subtracting the second equation from the first, we get:

$(a + (p-1)d) - (a + (q-1)d) = q - p$

Simplifying:

$pd - qd = q - p$

Factoring out $$d$$:

$d(p - q) = q - p$

Dividing both sides by $$(p - q)$$:

$d = -1$

Now that we know $$d = -1$$, we can substitute this back into either of the equations to find $$a$$. Let's use the first equation:

$a_p = a + (p-1)d = q$

$a + (p-1)(-1) = q$

$a - p + 1 = q$

Solving for $$a$$:

$a = q + p - 1$

So, the $$r$$th term is given by:

$a_r = q + p - 1 + (r-1)(-1)$

$a_r = q + p - r$

b) Find the sum of all the integers between 100 and 1000 that are divisible by 9:

To find the sum of all integers between 100 and 1000 that are divisible by 9, we can use the formula for the sum of an arithmetic series:

$S = \frac{n}{2} \times (a + l)$

where $$n$$ is the number of terms, $$a$$ is the first term, and $$l$$ is the last term.

The first term divisible by 9 in the given range is 108 (9  12), and the last term is 999 (9  111).

Let's find the number of terms:

$n = \frac{l - a}{d} + 1$

$n = \frac{999 - 108}{9} + 1$

$n = \frac{891}{9} + 1$

$n = 100$

Now, plug the values into the sum formula:

$S = \frac{100}{2} \times (108 + 999)$

$S = \frac{100}{2} \times 1107$

$S = \frac{110700}{2}$

$S = 55350$

So, the sum of all integers between 100 and 1000 that are divisible by 9 is 55350.