answer:1. **Extend ( PB ) to ( Q ) such that ( BQ = PA ).** - By construction, ( PQ = PB + BQ = PB + PA ). - We need to prove that ( D ) is the midpoint of ( PQ ), or equivalently, that ( CP = CQ ) since ( CD perp PQ ). 2. **Consider triangles ( triangle CAP ) and ( triangle CBQ ).** - We will show that these triangles are congruent. - Given ( CA = CB ) (since ( CA ) and ( CB ) are radii of the circumcircle and ( C ) is the midpoint of the arc ( AB )). - By construction, ( AP = BQ ). 3. **Angles ( angle CAP ) and ( angle CBQ ) are equal.** - Since ( CAPB ) is a cyclic quadrilateral, ( angle CAP = angle CBQ ). 4. **By SAS (Side-Angle-Side) congruence criterion:** - ( triangle CAP cong triangle CBQ ). - Therefore, ( CP = CQ ). 5. **Since ( CD perp PQ ) and ( CP = CQ ), ( D ) is the midpoint of ( PQ ).** - Hence, ( PD = frac{PQ}{2} = frac{PA + PB}{2} ). Thus, we have shown that ( PA + PB = 2 cdot PD ). (blacksquare) The final answer is ( boxed{ PA + PB = 2 cdot PD } ).

answer:To find all values of ( n ), where ( n in mathbb{N} ), such that the sum of the first ( n ) terms of the sequence ( a_k = 3k^2 - 3k + 1 ) is equal to the sum of the first ( n ) terms of the sequence ( b_k = 2k + 89 ), we proceed as follows: 1. **Sum of the Sequence ( a_k )**: - Notice that ( a_k = 3k^2 - 3k + 1 ). We recognize that: [ a_k = k^3 - (k-1)^3 ] This is due to the identity: [ k^3 - (k-1)^3 = (k-(k-1)) left(k^2 + k(k-1) + (k-1)^2 right) = 1 left(k^2 + k^2 - k + (k^2 - 2k + 1) right) = 3k^2 - 3k + 1 ] - Next, the sum of the first ( n ) terms of this sequence, using the property of telescoping series, is: [ S_n = sum_{k=1}^{n} (k^3 - (k-1)^3) = n^3 ] Here the intermediate terms cancel out, leaving just: [ S_n = n^3 ] 2. **Sum of the Sequence ( b_k )**: - ( b_k ) is given by ( b_k = 2k + 89 ). We can split ( b_k ) into simpler components: [ b_k = 2k + 89 = (k+45)^2 - (k+44)^2 ] This is derived using the identity: [ (k+45)^2 - (k+44)^2 = (k+45+k+44)(k+45-(k+44)) = (2k+89)(1) = 2k + 89 ] - Therefore, the sum of the first ( n ) terms of ( b_k ) becomes: [ S_n = sum_{k=1}^{n} ((k+45)^2 - (k+44)^2) = (n+45)^2 - 45^2 ] The intermediate terms cancel out, leaving us with: [ S_n = n(n+90) ] 3. **Setting the Sums Equal**: To find ( n ) where the sums are equal: [ n^3 = n(n+90) ] Dividing both sides by ( n ) (assuming ( n neq 0 )): [ n^2 = n + 90 ] Rearranging gives a quadratic equation: [ n^2 - n - 90 = 0 ] 4. **Solving the Quadratic Equation**: - To solve ( n^2 - n - 90 = 0 ), use the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -1 ), and ( c = -90 ): [ n = frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 1 cdot (-90)}}{2 cdot 1} = frac{1 pm sqrt{1 + 360}}{2} = frac{1 pm sqrt{361}}{2} ] [ n = frac{1 pm 19}{2} ] Hence, we get: [ n = frac{20}{2} = 10 quad text{or} quad n = frac{-18}{2} = -9 ] - Since ( n ) must be a natural number: [ n = 10 ] # Conclusion: The only natural number ( n ) that satisfies the given condition is: [ boxed{10} ]

answer:For n geq 3, we have [b_n = sqrt[(n-1)]{b_1 cdot b_2 cdot ldots cdot b_{n-1}}] or [(b_n)^{n-1} = b_1 cdot b_2 cdot ldots cdot b_{n-1}.] Similarly, [(b_{n+1})^n = b_1 cdot b_2 cdot ldots cdot b_{n-1} cdot b_n.] Dividing these two equations, we get [frac{(b_{n+1})^n}{(b_n)^{n-1}} = b_n,] hence [b_{n+1} = b_n sqrt[n]{frac{1}{b_n}} = sqrt[n]{b_n}.] This shows that from b_3 onwards, each term is simply the n-th root of the previous term. However, this seems to imply a declining series. Given b_9 = 45, then b_3 = b_9 = 45. Now, b_3 = sqrt[2]{b_1 b_2}, hence [45 = sqrt{10 cdot b_2}.] Squaring both sides gives [2025 = 10 cdot b_2,] which simplifies to [b_2 = frac{2025}{10} = 202.5.] Conclusion: Thus, the value of b_2 is boxed{202.5}.

answer:1. Divide both sides of the equation by 5: [ 2^y = frac{320}{5} = 64 ] 2. Recognize that 64 is 2^6. 3. Therefore, y = 6. Thus, the solution is: [ boxed{y = 6} ]