Example Question - sum formula
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Arithmetic Sequence Problem
<p>لإيجاد الحد الخامس ($a_5$) في متتالية حسابية حيث الحد الثامن هو $18$ ومجموع أول $8$ أحداث هو $72$، علينا أولا إيجاد الفرق الشائع ($d$).</p> <p>مجموع أول $n$ حدود في متتالية حسابية معطى بالعلاقة:</p> \[ S_n = \frac{n}{2}(2a_1 + (n-1)d) \] <p>حيث $S_n$ المجموع، $n$ عدد الحدود، $a_1$ الحد الأول، و$d$ الفرق الشائع. </p> <p>لدينا $S_8 = 72$ و $a_8 = 18$. $a_8$ معطى بالعلاقة:</p> \[ a_8 = a_1 + 7d \] <p>بالتعويض في معادلة المجموع:</p> \[ 72 = \frac{8}{2}(2a_1 + 7d) \] \[ 72 = 4(2a_1 + 7d) \] \[ 18 = 2a_1 + 7d \] \[ 9 = a_1 + 3.5d \] <p>لدينا نظام المعادلات:</p> \[ a_8 = a_1 + 7d = 18 \] \[ a_1 + 3.5d = 9 \] <p>نطرح المعادلة الثانية من الأولى:</p> \[ 7d - 3.5d = 18 - 9 \] \[ 3.5d = 9 \] \[ d = \frac{9}{3.5} = 2.571 \] <p>يمكن الآن حساب $a_1$ من المعادلة $a_1 + 3.5d = 9$:</p> \[ a_1 + 3.5 \cdot 2.571 = 9 \] \[ a_1 + 9 = 9 \] \[ a_1 = 0 \] <p>الآن يمكن حساب الحد الخامس:</p> \[ a_5 = a_1 + 4d \] \[ a_5 = 0 + 4 \cdot 2.571 \] \[ a_5 = 10.284 \] <p>إذًا، الحد الخامس هو تقريبًا $10.284$.</p>
Calculating the Sum of a Geometric Series
This question presents a finite geometric series that we need to sum up. A geometric series is given by a formula that involves the first term (a), the common ratio (r), and the number of terms (n). The sum (S) of the first n terms of a geometric series can be calculated with the formula: \[ S = a \cdot \frac{1 - r^n}{1 - r} \] for \( r ≠ 1 \). To use this formula, we must first find the values of a, r, and n for this series. The first term, \( a \), is clearly 0.5. The common ratio, \( r \), can be found by comparing any two successive terms. For instance, to get from the first term 0.5 to the next term 1.5, we multiply by 3 (since 0.5 * 3 = 1.5). We need to check that this same ratio applies to subsequent terms: \[ \frac{1.5}{0.5} = 3, \quad \frac{4.5}{1.5} = 3, \quad \text{etc.} \] Thus, the common ratio \( r \) is 3. To find \( n \), the number of terms in the series, we need to find how many times we have to multiply the first term by 3 to get to the last term given (88,573.5). Put it in the formula: \[ a \cdot r^{n-1} = \text{last term} \] Substitute the known values: \[ 0.5 \cdot 3^{n-1} = 88,573.5 \] Solve for \( n \) by first dividing both sides by 0.5: \[ 3^{n-1} = \frac{88,573.5}{0.5} \] \[ 3^{n-1} = 177,147 \] Now, knowing that \( 3^5 = 243 \) and \( 3^6 = 729 \), we aim to find the power of 3 that equals 177,147. Without checking further powers, it's evident that the number is quite large and this step might be impractical to do purely by hand, so this would typically be done using a logarithmic operation or an electronic calculator. Since I can't use an external calculator, let's simplify analytically: \[ 3^{n-1} = 177,147 \] Since \( 3^1 = 3 \), \( 3^2 = 9 \), \( 3^3 = 27 \), \( 3^4 = 81 \), and \( 3^5 = 243 \), it seems that \( 3^{11} \) would be close since \( 3^{10} \) is \( 59049 \). If we use the property \( 3^{11} = 3^{10} \times 3 \), it becomes clear that \( 3^{11} = 177,147 \) which implies that \( n-1 = 11 \), or \( n = 12 \). Now, substitute \( a = 0.5\), \( r = 3\), and \( n = 12 \) into the sum formula for the geometric series: \[ S = 0.5 \cdot \frac{1 - 3^{12}}{1 - 3} \] Since \( 1 - 3^{12} \) is \( 1 - 531,441 \) and \( 1 - 3 \) is \( -2\), the modified formula becomes: \[ S = 0.5 \cdot \frac{ - 531,440}{-2} \] \[ S = 0.5 \cdot 265,720 \] \[ S = 132,860 \] So the sum of the series is 132,860, which makes the correct answer B) 132,860.
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