What is 24th term of an arithmetic sequence if its 11th term is 29 and its 31st term is 169?
SAT Math questions from Sequences and Series are relatively easy ones.
Use the formula given in the 'Hint / Formula' tab and try and solve the question. If you need further assistance, click on the tabs provided to the right of the hint / formula tab for a detailed step wise explanation to solve the question.
The nth term of an arithmetic sequence t_{n} = t_{1} + (n - 1)d, where t_{1} is the first term of the arithmetic progression and 'd' is the common difference.
To find the 24th term of the arithmetic sequence, you will need the first term t_{1} and the common difference 'd'.
Data given: 11th term and 31st term of the same arithmetic sequence.
The 11th term of the sequence t_{11} = t_{1} + (11 - 1)*d
or t_{11} = 29 = t_{1} + 10d ..... eqn (1)
The 31st term of the sequence t_{31} = t_{1} + (31 - 1)d
or t_{31} = 169 = t_{1} + 30d ...... eqn (2)
Subtract eqn (1) from eqn (2)
t_{31} - t_{11} = 169 - 29 =140 = t_{1} + 30 d - (t_{1} + 10d)
or 140 = 20d
So, d = 7.
Substitute the value of d in eqn (1)
29 = t_{1} + 10*7
or t_{1} = 29 - 70 = -41.
t_{24} = t_{1} + (24 - 1)d
t_{24} = -41 + 23 * 7
t_{24} = -41 + 161
t_{24} = 120
The difference between any two consecutive terms of an arithmetic sequence is d. So, the difference between the 11th and the 31st term of an arithmetic sequence is 20 common differences.
t_{31} – t_{11} = 20d
169 – 29 = 140 = 20d
So, d= 7.
The difference between the 11th term and the 24th term is 13 common difference.
So, t_{24} = t_{11} + 13d
or t_{24} = 29 + 13*7
Hence, t_{24} = 29 + 91
t_{24} = 120
Get a quick overview on the basic concepts of Arithmetic sequence including formulae to find the nth term of an arithmetic sequence and the sum of n terms of an arithmetic sequence.
Concept Overview