💡Okay, so I have this inequality to prove in triangle ABC. It looks a bit complicated, but I'll try to break it down step by step. The inequality is:√(m² + tan(A/2) tan(B/2)) + √(n² + tan(B/2) tan(C/2)) + √(l² + tan(C/2) tan(A/2)) ≤ √[3(m² + n² + l² + 1)]Hmm, okay. So, it's an inequality involving square roots and tangent functions of half-angles in a triangle. I remember that in a triangle, the sum of angles is π, so each angle is between 0 and π, and their halves are between 0 and π/2, where the tangent function is positive. That might be useful.First, I need to recall some properties of triangles and trigonometric identities. Maybe I can use some known inequalities or identities related to triangle angles or sides. Since the inequality involves square roots, perhaps the Cauchy-Schwarz inequality could be helpful here. Or maybe the AM-QM inequality?Let me write down what I know:1. In any triangle ABC, A + B + C = π.2. The tangent of half-angles can be expressed in terms of the sides and the semiperimeter. Specifically, tan(A/2) = √[(s - B)(s - C)/(s(s - A))], where s is the semiperimeter. But I'm not sure if that's directly helpful here.3. Maybe I can use the identity that in a triangle, tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2) = 1. Wait, is that true? Let me check.Actually, in a triangle, tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2) = 1. Yes, that's a known identity. So, that might be useful.So, the sum of the products of the tangents of half-angles is 1. That could be a key point.Now, looking back at the inequality, each term inside the square roots has m², n², l², and the product of tangents. So, maybe I can consider each square root term as a vector component and then apply some vector inequality.Alternatively, maybe I can use the Cauchy-Schwarz inequality on the sum of square roots.Let me recall the Cauchy-Schwarz inequality: For non-negative real numbers a_i and b_i, (Σa_i b_i)² ≤ (Σa_i²)(Σb_i²). But in this case, I have a sum of square roots, which might not directly fit into that form.Wait, another approach: Maybe I can use the QM-AM inequality, which states that the quadratic mean is greater than or equal to the arithmetic mean. But in this case, I have a sum of square roots on the left and a single square root on the right. So, perhaps squaring both sides would help.Yes, squaring both sides is a common technique when dealing with square roots. Let me try that.Let me denote the left-hand side as L and the right-hand side as R:L = √(m² + tan(A/2) tan(B/2)) + √(n² + tan(B/2) tan(C/2)) + √(l² + tan(C/2) tan(A/2))R = √[3(m² + n² + l² + 1)]So, squaring both sides:L² = [√(m² + tan(A/2) tan(B/2)) + √(n² + tan(B/2) tan(C/2)) + √(l² + tan(C/2) tan(A/2))]²R² = 3(m² + n² + l² + 1)Expanding L², we get:L² = (m² + tan(A/2) tan(B/2)) + (n² + tan(B/2) tan(C/2)) + (l² + tan(C/2) tan(A/2)) + 2[√{(m² + tan(A/2) tan(B/2))(n² + tan(B/2) tan(C/2))} + √{(n² + tan(B/2) tan(C/2))(l² + tan(C/2) tan(A/2))} + √{(l² + tan(C/2) tan(A/2))(m² + tan(A/2) tan(B/2))}]Simplifying, L² becomes:L² = m² + n² + l² + [tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2)] + 2[√{(m² + tan(A/2) tan(B/2))(n² + tan(B/2) tan(C/2))} + √{(n² + tan(B/2) tan(C/2))(l² + tan(C/2) tan(A/2))} + √{(l² + tan(C/2) tan(A/2))(m² + tan(A/2) tan(B/2))}]But from the identity I recalled earlier, tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2) = 1. So, substituting that in:L² = m² + n² + l² + 1 + 2[√{(m² + tan(A/2) tan(B/2))(n² + tan(B/2) tan(C/2))} + √{(n² + tan(B/2) tan(C/2))(l² + tan(C/2) tan(A/2))} + √{(l² + tan(C/2) tan(A/2))(m² + tan(A/2) tan(B/2))}]So, L² = (m² + n² + l² + 1) + 2[sum of cross terms]Now, R² = 3(m² + n² + l² + 1)So, to prove that L² ≤ R², we need to show that:(m² + n² + l² + 1) + 2[sum of cross terms] ≤ 3(m² + n² + l² + 1)Subtracting (m² + n² + l² + 1) from both sides:2[sum of cross terms] ≤ 2(m² + n² + l² + 1)Dividing both sides by 2:sum of cross terms ≤ (m² + n² + l² + 1)So, now the problem reduces to showing that:√{(m² + tan(A/2) tan(B/2))(n² + tan(B/2) tan(C/2))} + √{(n² + tan(B/2) tan(C/2))(l² + tan(C/2) tan(A/2))} + √{(l² + tan(C/2) tan(A/2))(m² + tan(A/2) tan(B/2))} ≤ (m² + n² + l² + 1)Hmm, okay. So, I need to bound the sum of these square roots by (m² + n² + l² + 1). Maybe I can use the AM-GM inequality here.Recall that for non-negative real numbers a and b, √(ab) ≤ (a + b)/2. So, maybe applying this to each cross term.Let me try that.First term: √{(m² + tan(A/2) tan(B/2))(n² + tan(B/2) tan(C/2))} ≤ [ (m² + tan(A/2) tan(B/2)) + (n² + tan(B/2) tan(C/2)) ] / 2Similarly for the other terms.So, sum of cross terms ≤ [ (m² + tan(A/2) tan(B/2)) + (n² + tan(B/2) tan(C/2)) ] / 2 + [ (n² + tan(B/2) tan(C/2)) + (l² + tan(C/2) tan(A/2)) ] / 2 + [ (l² + tan(C/2) tan(A/2)) + (m² + tan(A/2) tan(B/2)) ] / 2Simplifying the right-hand side:Each pair of terms is added twice, so:= [2(m² + n² + l²) + 2(tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2)) ] / 2= (m² + n² + l²) + (tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2))But we know that tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2) = 1, so:= (m² + n² + l²) + 1Therefore, sum of cross terms ≤ (m² + n² + l²) + 1Which is exactly what we needed to show. Therefore, L² ≤ R², which implies L ≤ R.So, the original inequality holds.Wait, let me double-check my steps. I used the AM-GM inequality on each cross term, which is valid because all terms are non-negative. Then, I summed them up and found that the sum is bounded by (m² + n² + l² + 1). That seems correct.Also, I used the identity that the sum of the products of the tangents of half-angles is 1, which is a known identity in triangle geometry. So, that step is also correct.Therefore, the proof seems solid. I think I've got it!

💡First, I need to identify the tenths place in the number 12345.64789. The tenths place is the first digit after the decimal point, which is 6.Next, I look at the hundredths place to determine whether to round up or down. The hundredths place is the second digit after the decimal point, which is 4.Since the hundredths place is less than 5, I do not round up the tenths place. Therefore, the tenths place remains 6.Finally, I drop all the digits after the tenths place, resulting in the number 12345.6.

💡Okay, so I have this problem where I'm given a function ( f(x) = x^2 + 2x + a ln x ). The condition is that for any ( t geq 1 ), the inequality ( f(2t - 1) geq 2f(t) - 3 ) holds. I need to find the range of values for the real number ( a ).Alright, let me start by writing down the given inequality:[ f(2t - 1) geq 2f(t) - 3 ]Since ( f(x) = x^2 + 2x + a ln x ), I can substitute this into the inequality:[ (2t - 1)^2 + 2(2t - 1) + a ln(2t - 1) geq 2(t^2 + 2t + a ln t) - 3 ]Let me expand both sides step by step.First, expand the left-hand side (LHS):1. ( (2t - 1)^2 = 4t^2 - 4t + 1 )2. ( 2(2t - 1) = 4t - 2 )3. So, combining these, the LHS becomes: [ 4t^2 - 4t + 1 + 4t - 2 + a ln(2t - 1) ] Simplify: [ 4t^2 - 4t + 4t + 1 - 2 + a ln(2t - 1) ] Which simplifies to: [ 4t^2 - 1 + a ln(2t - 1) ]Now, expand the right-hand side (RHS):1. ( 2(t^2 + 2t + a ln t) = 2t^2 + 4t + 2a ln t )2. Subtract 3: [ 2t^2 + 4t + 2a ln t - 3 ]So, putting it all together, the inequality becomes:[ 4t^2 - 1 + a ln(2t - 1) geq 2t^2 + 4t + 2a ln t - 3 ]Let me bring all terms to the left-hand side to simplify:[ 4t^2 - 1 + a ln(2t - 1) - 2t^2 - 4t - 2a ln t + 3 geq 0 ]Simplify term by term:1. ( 4t^2 - 2t^2 = 2t^2 )2. ( -4t ) remains as is.3. ( -1 + 3 = 2 )4. ( a ln(2t - 1) - 2a ln t = a [ln(2t - 1) - 2 ln t] )So, the inequality becomes:[ 2t^2 - 4t + 2 + a [ln(2t - 1) - 2 ln t] geq 0 ]Hmm, that's a bit complicated. Maybe I can factor out some terms or rewrite the logarithmic part.Let me look at the logarithmic term:[ ln(2t - 1) - 2 ln t = ln(2t - 1) - ln t^2 = lnleft( frac{2t - 1}{t^2} right) ]So, substituting back, the inequality becomes:[ 2t^2 - 4t + 2 + a lnleft( frac{2t - 1}{t^2} right) geq 0 ]Hmm, maybe I can factor the quadratic part:[ 2t^2 - 4t + 2 = 2(t^2 - 2t + 1) = 2(t - 1)^2 ]So, now the inequality is:[ 2(t - 1)^2 + a lnleft( frac{2t - 1}{t^2} right) geq 0 ]Let me denote this expression as ( u(t) ):[ u(t) = 2(t - 1)^2 + a lnleft( frac{2t - 1}{t^2} right) ]So, the condition is ( u(t) geq 0 ) for all ( t geq 1 ).To analyze this, I might need to look at the behavior of ( u(t) ) as ( t ) increases and check its minimum value.First, let's check the value at ( t = 1 ):[ u(1) = 2(0)^2 + a lnleft( frac{2(1) - 1}{1^2} right) = 0 + a ln(1) = 0 ]So, at ( t = 1 ), ( u(t) = 0 ). That's the boundary.Now, let's compute the derivative ( u'(t) ) to see if ( u(t) ) is increasing or decreasing around ( t = 1 ) and beyond.Compute ( u'(t) ):[ u(t) = 2(t - 1)^2 + a lnleft( frac{2t - 1}{t^2} right) ]Differentiate term by term:1. The derivative of ( 2(t - 1)^2 ) is ( 4(t - 1) ).2. The derivative of ( a lnleft( frac{2t - 1}{t^2} right) ) is ( a cdot frac{d}{dt} left[ ln(2t - 1) - 2 ln t right] ).Compute the derivative inside the logarithm:[ frac{d}{dt} ln(2t - 1) = frac{2}{2t - 1} ][ frac{d}{dt} (-2 ln t) = -frac{2}{t} ]So, combining these:[ frac{d}{dt} lnleft( frac{2t - 1}{t^2} right) = frac{2}{2t - 1} - frac{2}{t} ]Therefore, the derivative of the logarithmic term is:[ a left( frac{2}{2t - 1} - frac{2}{t} right) ]Putting it all together, the derivative ( u'(t) ) is:[ u'(t) = 4(t - 1) + a left( frac{2}{2t - 1} - frac{2}{t} right) ]Simplify the expression:First, factor out the 2 in the logarithmic derivative:[ u'(t) = 4(t - 1) + 2a left( frac{1}{2t - 1} - frac{1}{t} right) ]Combine the fractions:[ frac{1}{2t - 1} - frac{1}{t} = frac{t - (2t - 1)}{t(2t - 1)} = frac{t - 2t + 1}{t(2t - 1)} = frac{-t + 1}{t(2t - 1)} ]So, substitute back:[ u'(t) = 4(t - 1) + 2a cdot frac{ - (t - 1) }{ t(2t - 1) } ]Factor out ( (t - 1) ):[ u'(t) = (t - 1) left[ 4 - frac{2a}{t(2t - 1)} right] ]So, ( u'(t) = (t - 1) left( 4 - frac{2a}{t(2t - 1)} right) )Now, since ( t geq 1 ), ( t - 1 geq 0 ). Therefore, the sign of ( u'(t) ) depends on the term in the brackets:[ 4 - frac{2a}{t(2t - 1)} ]Let me denote this term as ( v(t) = 4 - frac{2a}{t(2t - 1)} )So, ( u'(t) = (t - 1) v(t) )We need to determine when ( u'(t) geq 0 ) for ( t geq 1 ). Since ( t - 1 geq 0 ), this is equivalent to ( v(t) geq 0 ).So, ( v(t) = 4 - frac{2a}{t(2t - 1)} geq 0 )Let me solve for ( a ):[ 4 geq frac{2a}{t(2t - 1)} ][ 4 t(2t - 1) geq 2a ][ 2 t(2t - 1) geq a ]So, ( a leq 2 t(2t - 1) ) for all ( t geq 1 ).But since this must hold for all ( t geq 1 ), the maximum lower bound for ( a ) is determined by the minimum value of ( 2 t(2t - 1) ) over ( t geq 1 ).Let me compute ( 2 t(2t - 1) ) at ( t = 1 ):[ 2(1)(2(1) - 1) = 2(1)(1) = 2 ]Now, as ( t ) increases beyond 1, ( 2 t(2t - 1) ) increases because it's a quadratic function opening upwards. So, the minimum value of ( 2 t(2t - 1) ) for ( t geq 1 ) is 2, occurring at ( t = 1 ).Therefore, ( a leq 2 ) is required for ( v(t) geq 0 ) for all ( t geq 1 ).But wait, let me verify this. If ( a leq 2 ), then ( v(t) geq 0 ) for all ( t geq 1 ), which means ( u'(t) geq 0 ) for all ( t geq 1 ). Therefore, ( u(t) ) is non-decreasing for ( t geq 1 ).Since ( u(1) = 0 ) and ( u(t) ) is non-decreasing, ( u(t) geq 0 ) for all ( t geq 1 ). That satisfies the original condition.But what if ( a > 2 )? Let's test ( a = 3 ) as an example.At ( a = 3 ), ( v(t) = 4 - frac{6}{t(2t - 1)} ). Let's see if ( v(t) ) is always non-negative.At ( t = 1 ):[ v(1) = 4 - frac{6}{1(2 - 1)} = 4 - 6 = -2 ]So, ( v(1) = -2 < 0 ). Therefore, ( u'(1) = (1 - 1)(...) = 0 ), but for ( t ) just above 1, ( u'(t) ) would be negative because ( v(t) ) is negative. Therefore, ( u(t) ) would decrease just after ( t = 1 ), making ( u(t) < 0 ) for some ( t > 1 ). Hence, the condition ( u(t) geq 0 ) is violated.Therefore, ( a ) must be less than or equal to 2.Wait, but let me check another point. Suppose ( a = 2 ). Then ( v(t) = 4 - frac{4}{t(2t - 1)} ).At ( t = 1 ):[ v(1) = 4 - frac{4}{1(2 - 1)} = 4 - 4 = 0 ]For ( t > 1 ), since ( t(2t - 1) ) increases, ( frac{4}{t(2t - 1)} ) decreases, so ( v(t) = 4 - text{something decreasing} ) increases. Therefore, ( v(t) geq 0 ) for all ( t geq 1 ).Thus, ( u'(t) geq 0 ) for all ( t geq 1 ), meaning ( u(t) ) is non-decreasing, starting from ( u(1) = 0 ). Therefore, ( u(t) geq 0 ) for all ( t geq 1 ).Therefore, the maximum value ( a ) can take is 2. If ( a ) is less than 2, ( v(t) ) will be even larger, so ( u'(t) ) will still be non-negative, ensuring ( u(t) geq 0 ).Hence, the range of ( a ) is all real numbers less than or equal to 2.**Final Answer**The range of values for ( a ) is boxed{(-infty, 2]}.**Final Answer**The range of values for the real number ( a ) is boxed{(-infty, 2]}.

💡First, I need to solve the inequality (102n^2 leq 8100) to find the greatest possible integer value of (n).I'll start by isolating (n^2) by dividing both sides of the inequality by 102:[n^2 leq frac{8100}{102}]Calculating the division:[n^2 leq 79.4117647]Next, I'll take the square root of both sides to solve for (n):[n leq sqrt{79.4117647} approx 8.91]Since (n) must be an integer, the greatest integer less than or equal to 8.91 is 8.Therefore, the greatest possible value of (n) is 8.

💡Alright, so I've got this problem here: In a five-digit number, one of the digits was crossed out, and the resulting four-digit number was subtracted from the original number. The result was 54321. I need to find the original number. Hmm, okay, let's break this down step by step.First, let's understand the problem. We have a five-digit number. Let's call this number ( n ). One of its digits is crossed out, which means we're removing one digit from it, resulting in a four-digit number. Let's call this four-digit number ( x ). Now, according to the problem, when we subtract this four-digit number ( x ) from the original five-digit number ( n ), we get 54321. So, mathematically, that can be written as:[n - x = 54321]Okay, so that's our main equation. Now, the challenge is to find ( n ), the original five-digit number.Let me think about how crossing out a digit affects the number. If I have a five-digit number, say ( ABCDE ), and I cross out one digit, say ( C ), then the resulting four-digit number would be ( ABDE ). So, essentially, the position of the crossed-out digit affects the value of the four-digit number.But how can I represent this mathematically? Well, if I consider the five-digit number ( n ) as:[n = 10000a + 1000b + 100c + 10d + e]where ( a, b, c, d, e ) are the digits of the number, with ( a ) being the ten-thousands place, ( b ) the thousands place, and so on.Now, if I cross out a digit, say ( c ), the resulting four-digit number ( x ) would be:[x = 1000a + 100b + 10d + e]So, in this case, the five-digit number ( n ) can be expressed in terms of ( x ) and the crossed-out digit ( c ). Specifically, ( n ) is equal to:[n = 10x + c]Wait, let me explain that. If ( x ) is the four-digit number obtained by removing the digit ( c ) from ( n ), then ( n ) can be thought of as ( x ) multiplied by 10 plus the digit ( c ). That makes sense because when you remove a digit from the middle of a number, you're effectively shifting the subsequent digits one place to the left, which is equivalent to multiplying by 10 and then adding the removed digit.So, substituting ( n = 10x + c ) into our original equation ( n - x = 54321 ), we get:[10x + c - x = 54321]Simplifying that:[9x + c = 54321]Okay, so now we have an equation involving ( x ) and ( c ). Since ( c ) is a single digit, it must be between 0 and 9. Therefore, ( c ) can take any value from 0 to 9. Our goal is to find integers ( x ) and ( c ) such that ( 9x + c = 54321 ).Let's rearrange the equation to solve for ( x ):[9x = 54321 - c][x = frac{54321 - c}{9}]Since ( x ) must be an integer (it's a four-digit number), ( 54321 - c ) must be divisible by 9. That means ( 54321 - c ) is a multiple of 9. So, ( c ) must be such that when subtracted from 54321, the result is divisible by 9.I remember that a number is divisible by 9 if the sum of its digits is divisible by 9. Let's check the sum of the digits of 54321:5 + 4 + 3 + 2 + 1 = 1515 is not divisible by 9, so 54321 is not divisible by 9. Therefore, ( 54321 - c ) must be the next number that is divisible by 9. To find ( c ), we can calculate the remainder when 54321 is divided by 9 and then determine what ( c ) needs to be to make the entire expression divisible by 9.Let's compute ( 54321 div 9 ):9 goes into 54 six times (54), remainder 0.Bring down the 3: 03. 9 goes into 3 zero times, remainder 3.Bring down the 2: 32. 9 goes into 32 three times (27), remainder 5.Bring down the 1: 51. 9 goes into 51 five times (45), remainder 6.So, 54321 divided by 9 is 6035 with a remainder of 6. Therefore,[54321 = 9 times 6035 + 6]This means that ( 54321 ) is 6 more than a multiple of 9. To make ( 54321 - c ) divisible by 9, ( c ) must be equal to 6. Because subtracting 6 from 54321 will give us a multiple of 9:[54321 - 6 = 54315]Now, let's check if 54315 is divisible by 9:5 + 4 + 3 + 1 + 5 = 1818 is divisible by 9, so yes, 54315 is divisible by 9.Therefore, ( c = 6 ), and:[x = frac{54321 - 6}{9} = frac{54315}{9} = 6035]So, ( x = 6035 ) and ( c = 6 ).Now, recalling that ( n = 10x + c ), we can substitute the values of ( x ) and ( c ):[n = 10 times 6035 + 6 = 60350 + 6 = 60356]Therefore, the original five-digit number is 60356.But wait, let me double-check to make sure I didn't make any mistakes. Let's verify the result:Original number: 60356Cross out the digit 6 (which is the thousands place digit):Resulting four-digit number: 0356. Wait, that's not right. If we cross out the digit 6 in 60356, we get 0356, but leading zeros are not considered in numbers, so it would actually be 356, which is a three-digit number, not four-digit. Hmm, that's a problem.Wait, maybe I crossed out the wrong digit. Let me think again. The original number is 60356. If I cross out the digit 0, which is in the thousands place, then the resulting number would be 6356, which is a four-digit number. Let's check:60356 - 6356 = ?Calculating:60356- 6356--------54000Wait, that's not 54321. Hmm, that's not matching the result given in the problem.Wait, maybe I crossed out a different digit. Let's try crossing out the digit 3:Original number: 60356Cross out the 3: 6056Wait, that's a four-digit number. Now, subtracting:60356 - 6056 = ?60356- 6056--------54300Hmm, that's 54300, which is close to 54321 but not exactly.Wait, maybe I crossed out the digit 5:Original number: 60356Cross out the 5: 6036Subtracting:60356 - 6036 = ?60356- 6036--------54320Oh, that's 54320, which is just one less than 54321. Hmm, almost there.Wait, maybe I crossed out the digit 6 at the end:Original number: 60356Cross out the last 6: 6035Subtracting:60356 - 6035 = ?60356- 6035--------54321Ah, there we go! That gives us exactly 54321.So, in this case, crossing out the last digit 6 from 60356 gives us 6035, and subtracting 6035 from 60356 gives us 54321, which matches the problem statement.Therefore, my initial calculation was correct, but I had to make sure which digit was being crossed out to get the correct four-digit number. It turns out that crossing out the last digit 6 gives the correct result.So, the original number is indeed 60356.

💡Okay, so I have this problem about an arithmetic sequence. Let me try to understand what it's asking. The sequence is arithmetic, which means each term increases by a constant difference. They told me the common difference is 2, so that's good to know. The problem also says that the first term, a₁, the third term, a₃, and the fourth term, a₄, form a geometric sequence. Hmm, okay. So, in a geometric sequence, each term is multiplied by a common ratio to get the next term. That means the ratio between a₃ and a₁ should be the same as the ratio between a₄ and a₃. Let me write down what I know. In an arithmetic sequence, each term is the previous term plus the common difference. So, a₁ is just a₁, a₂ is a₁ + 2, a₃ is a₁ + 4, and a₄ is a₁ + 6. Wait, let me check that. If the common difference is 2, then:- a₁ = a₁- a₂ = a₁ + 2- a₃ = a₂ + 2 = a₁ + 4- a₄ = a₃ + 2 = a₁ + 6Yes, that seems right. So, a₁, a₃, and a₄ are a₁, a₁ + 4, and a₁ + 6 respectively.Now, these three terms form a geometric sequence. In a geometric sequence, the ratio between consecutive terms is constant. So, the ratio of a₃ to a₁ should be equal to the ratio of a₄ to a₃. Let me write that as an equation:(a₃) / (a₁) = (a₄) / (a₃)Substituting the expressions I have for a₃ and a₄:(a₁ + 4) / a₁ = (a₁ + 6) / (a₁ + 4)Okay, so now I have an equation with a₁. I need to solve for a₁. Let me cross-multiply to get rid of the fractions:(a₁ + 4)^2 = a₁ * (a₁ + 6)Expanding both sides:Left side: (a₁ + 4)^2 = a₁^2 + 8a₁ + 16Right side: a₁*(a₁ + 6) = a₁^2 + 6a₁So, putting it all together:a₁^2 + 8a₁ + 16 = a₁^2 + 6a₁Hmm, let me subtract a₁^2 from both sides to simplify:8a₁ + 16 = 6a₁Now, subtract 6a₁ from both sides:2a₁ + 16 = 0Subtract 16 from both sides:2a₁ = -16Divide both sides by 2:a₁ = -8Wait, so a₁ is -8? Let me check if this makes sense. If a₁ is -8, then:a₁ = -8a₃ = a₁ + 4 = -8 + 4 = -4a₄ = a₁ + 6 = -8 + 6 = -2So, the terms are -8, -4, -2. Is that a geometric sequence? Let's see the ratios:-4 / -8 = 0.5-2 / -4 = 0.5Yes, both ratios are 0.5, so it is a geometric sequence with a common ratio of 1/2. That checks out.Wait, but let me make sure I didn't make any mistakes in my algebra. Let me go through the steps again.Starting with (a₁ + 4)^2 = a₁*(a₁ + 6)Expanding left side: a₁² + 8a₁ + 16Right side: a₁² + 6a₁Subtract right side from left side: (a₁² + 8a₁ + 16) - (a₁² + 6a₁) = 0Simplify: 2a₁ + 16 = 0So, 2a₁ = -16 => a₁ = -8Yes, that seems correct. So, a₁ must be -8. Looking at the options, that's option C.I think that's it. I don't see any mistakes in my calculations, and the check confirms that the terms do form a geometric sequence. So, the answer should be -8.**Final Answer**boxed{C}