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## SAT Arithmetic Progression : Sum of select terms of an AP

### Question 5: Sum of 41st to 49th term of an Arithmetic Sequence

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?

1. 4104
2. 1026
3. 3852
4. 456
5. 2052
Choice E. Sum of the 41st to 49th term is 2052.

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.

1. Find the common difference of the arithmetic sequence using the data given.
2. Use this information to find the 41st term of the sequence.
3. The sum of the sequence from the 41st term to the 49th term is the sum of 9 consecutive terms of the sequence starting from the 41st term.

### Alternative Method

1. Find out which term will be the average of the first 9 terms of this sequence and the value of the average.
2. Use this information in combination with the first term to determine the common difference.
3. Find the term that will be the average of the 41st to the 49th term and compute it.
4. Use this information to find the sum.

### Step 1: Compute the common difference of the arithmetic sequence

The following information is available

1. First term of the arithmetic sequence is 8.
2. The sum of the first 9 terms is 252.

Sum of first n terms of an arithmetic sequence Sn = $n $over 2 \\$ $\ \left\lbrack 2t_1 + \lbrack n-1 \rbrack d \right\rbrack \\$ Sum of first n terms of an arithmetic sequence S9 = $9 \over 2 \\$ $\ \left\lbrack 2t_1 + \lbrack 9-1 \rbrack d \right\rbrack \\$ S9 = 252 = $9 \over 2 \\$ $\ \left\lbrack 2*8 + 8d \right\rbrack \\$ The commond difference d = 5. ### Step 2: Compute the 41st term of the arithmetic sequence $\ t_{41} = t_1 + \lbrack 41 - 1 \rbrack d \\$ $\ t_{41} = t_1 + 40 * d \\$ $\ t_{41} = 8 + 40 * 5 \\$ $\ t_{41} = 208 \\$ ### Step 3: Compute the sum of the 41st to 49th term of the sequence 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. $\ S_{41\;to\; 49} \\$ = $9 \over 2 \\$ $\ \left\lbrack 2t_{41} + \lbrack 9-1 \rbrack d \right\rbrack \\$ $\ S_{41\;to\; 49} \\$ = $9 \over 2 \\$ $\ \left\lbrack 2*208 + 8 * 5 \right\rbrack \\$ $\ S_{41\;to\; 49} \\$ = $9 \over 2 \\$ $\ \left\lbrack 456 \right\rbrack \\$ S41 to 49 = 2052 ### Alternative Method 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 t5 = 28

t1 = 8; t5 = 28.

t5 - t1 = 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.

t45 = t1 + 44d = 8 + 44*5 = 228

Or S41 to 49 = 9 * 228 = 2052

## Concept Overview : Arithmetic Progression

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.

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