In an arithmetic progression (t_3+t_9=70) and (t_5+t_{15}=110). What is the sum of the first (20) terms?
Answer and explanation
Correct answer: 575
Let the first term be \(a\) and the common difference be \(d\). From \(t_3+t_9=(a+2d)+(a+8d)=70\), we get \(a+5d=35\). Similarly, \(t_5+t_{15}=(a+4d)+(a+14d)=110\) gives \(a+9d=55\). Subtracting gives \(4d=20\), so \(d=5\) and \(a=10\). Therefore, \(S_{20}=\frac{20}{2}[2(10)+19(5)]=575\). The value 1150 can result from incorrectly using \(n\) instead of \(\frac{n}{2}\) in the sum formula. Exam tip: convert the given term relations into equations in \(a\) and \(d\) first.
Frequently asked questions
What is the correct answer to this question?
575
Why is this the correct answer?
Let the first term be \(a\) and the common difference be \(d\). From \(t_3+t_9=(a+2d)+(a+8d)=70\), we get \(a+5d=35\). Similarly, \(t_5+t_{15}=(a+4d)+(a+14d)=110\) gives \(a+9d=55\). Subtracting gives \(4d=20\), so \(d=5\) and \(a=10\). Therefore, \(S_{20}=\frac{20}{2}[2(10)+19(5)]=575\). The value 1150 can result from incorrectly using \(n\) instead of \(\frac{n}{2}\) in the sum formula. Exam tip: convert the given term relations into equations in \(a\) and \(d\) first.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.
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