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.
0 टिप्पणियाँ:
Post a Comment