In a linear sequence, \(a_5=23\) and \(a_{12}=58\). What will be \(a_{20}\)?
Answer and explanation
Correct answer: 98
A linear sequence has a constant first difference, so write \(a_n=dn+c\). From the fifth to the twelfth term, the index increases by 7 while the value increases by \(58-23=35\). Hence \(7d=35\), giving \(d=5\). Moving from the 12th term to the 20th term adds 8 steps, so \(a_{20}=58+8(5)=98\). Equivalently, using \(a_n=5n-2\), we get \(a_{20}=100-2=98\). Thus option A is correct. The other choices arise from using the wrong number of steps or an arithmetic error.
Frequently asked questions
What is the correct answer to this question?
98
Why is this the correct answer?
A linear sequence has a constant first difference, so write \(a_n=dn+c\). From the fifth to the twelfth term, the index increases by 7 while the value increases by \(58-23=35\). Hence \(7d=35\), giving \(d=5\). Moving from the 12th term to the 20th term adds 8 steps, so \(a_{20}=58+8(5)=98\). Equivalently, using \(a_n=5n-2\), we get \(a_{20}=100-2=98\). Thus option A is correct. The other choices arise from using the wrong number of steps or an arithmetic error.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.
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