The first term of an arithmetic sequence is 8 and the sum of its first 9 terms is 252. What is the sum of the 41st to the 49th term of the arithmetic sequence?
An interesting arithmetic progression question. We have to find out the sum of a specific set of terms of an arithmetic sequence - not the first 49 terms as is the usual case - but the sum of the 41st to the 49th term of the arithmetic progression. The level of difficulty is moderate. Use the Hints/Clue given in the first tab to solve the question. If you need more help with this SAT practice question, click on the subsequent tabs to get a detailed explanation.
The following information is available
Sum of first n terms of an arithmetic sequence S_{n} = \n \over 2 \\) \\ \left\lbrack 2t_1 + \lbrack n-1 \rbrack d \right\rbrack \\)
Sum of first n terms of an arithmetic sequence S_{9} = \9 \over 2 \\) \\ \left\lbrack 2t_1 + \lbrack 9-1 \rbrack d \right\rbrack \\)
S_{9} = 252 = \9 \over 2 \\) \\ \left\lbrack 2*8 + 8d \right\rbrack \\)
The commond difference d = 5.
\\ t_{41} = t_1 + \lbrack 41 - 1 \rbrack d \\)
\\ t_{41} = t_1 + 40 * d \\)
\\ t_{41} = 8 + 40 * 5 \\)
\\ t_{41} = 208 \\)
The sum of the 41st to the 49th term of the arithmetic sequence is the sum of 9 terms of the arithmetic sequence starting with the 41st term and having a common difference of 5.
\\ S_{41\;to\; 49} \\) = \9 \over 2 \\) \\ \left\lbrack 2t_{41} + \lbrack 9-1 \rbrack d \right\rbrack \\)
\\ S_{41\;to\; 49} \\) = \9 \over 2 \\) \\ \left\lbrack 2*208 + 8 * 5 \right\rbrack \\)
\\ S_{41\;to\; 49} \\) = \9 \over 2 \\) \\ \left\lbrack 456 \right\rbrack \\)
S_{41 to 49} = 2052
The middle term of an arithmetic sequence is the average of the arithmetic sequence. (Refer to Concept Overview given below for details)
The average of the first 9 terms is the 5th term of the sequence.
The sum of the first 9 terms = 252.
So, average of the first 9 terms = \252 \over 9 \\) = 28
Therefore, the fifth term t_{5} = 28
t_{1} = 8; t_{5} = 28.
t_{5} - t_{1} = 4d = 28 – 8 = 20, where d is the common difference of the sequence
Or d = 5
Sum of the 9 terms from the 41st term to the 49th term = 9 * average of the 9 terms
The average of the 9 terms from the 41st term to the 49th term is the middle term = 45th term.
t_{45} = t_{1} + 44d = 8 + 44*5 = 228
Or S41 to 49 = 9 * 228 = 2052
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