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.

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