Let \(O\) be the centre of the circle
Given that,
\(OQ = 25\) cm and \(PQ = 24\)cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, \(OP ⊥ PQ\)
Applying Pythagoras theorem in \(\text {ΔOPQ}\), we obtain
\(OP^2 + PQ^2 = OQ^2\)
\(OP^2 + 24^2 = 25^2\)
\(OP^2 = 625 - 576\)
\(OP^2 = 49\)
\(OP = 7\)
Therefore, the radius of the circle is \(7\) cm.
Hence, the correct option is (A): \(7\) cm
Assertion (A): The sum of the first fifteen terms of the AP $ 21, 18, 15, 12, \dots $ is zero.
Reason (R): The sum of the first $ n $ terms of an AP with first term $ a $ and common difference $ d $ is given by: $ S_n = \frac{n}{2} \left[ a + (n - 1) d \right]. $
Assertion (A): The sum of the first fifteen terms of the AP $21, 18, 15, 12, \dots$ is zero.
Reason (R): The sum of the first $n$ terms of an AP with first term $a$ and common difference $d$ is given by: $S_n = \frac{n}{2} \left[ a + (n - 1) d \right].$