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In the sequence a_1, a_2, . . . , a_n each term

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by Gmat_mission » Fri Jun 01, 2018 1:20 am
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In the sequence a_1, a_2, . . . , a_n each term after the first term is equal to the preceding term plus a constant c. $$a_1+a_{11}+a_{21}=99.$$ What is the value of $$a_3+a_{19}=?$$ A. 66
B. 44
C. 33
D. 22
E. 11

[spoiler]OA=A[/spoiler].

What is the best way to solve this PS question? Can anyone give me some help, please? I'd be thankful.
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by [email protected] » Fri Jun 01, 2018 10:08 am
Hi Gmat_mission,

We're told that In the sequence a_1, a_2, . . . , a_n each term after the first term is equal to the preceding term plus a constant C and that $$a_1+a_{11}+a_{21}=99.$$ We're asked for the value of $$a_3+a_{19}=?$$.

Since we're dealing with a sequence that increases by a constant value (from term to term), it might help to create some examples so that you can get a better sense of the 'math' involved. Based on that general type of sequence, the sequence could be....
1, 2, 3, 4, 5.... this sequence begins with a 1 and increases by a constant (C=1) with each term.
2, 5, 8, 11, 14... this sequence begins with a 2 and increases by a constant (C=3) with each term.
Etc.

The additional information we're given tells us that the (1st term) + (11th term) + (21st term) = 99, so we can create an equation to solve for the potential values of the first term and the constant.
1st term = X
11th term = X + 10(C)
21st term = X + 20(C)
(X) + (X + 10C) + (X + 20C) = 99
3X + 30C = 99
X + 10C = 33

IF... X = C = 3, then....
1st term = 3
11th term = 33
21st term = 63

The value of the....
3rd term = 3 + 6 = 9
19th term = 3 + 54 = 57
And the answer to the question is 9+57 = 66

Final Answer: A

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by deloitte247 » Sat Jun 02, 2018 6:30 pm
Each term is greater than the previous term with an added C. Each term providing terms $$=a_x=\left(x+1\right)C$$ more than $$a_1$$
$$a_2=a_1+C$$

$$a_3=a_2+C$$ $$=\left(a_1+C\right)+C=a_1+2C$$

$$a_{_4}=\left(a_3+C\right)=a_1+3C$$

$$a_{_{11}}=\left(a_{10}+C\right)=a_1+10C$$

$$a_{_{21}}=\left(a_{20}+C\right)=a_1+20C$$
$$a_1+a_{11}+a_{21}=99$$
$$a_1+\left(a_1+10C\right)+\left(a_1+20C\right)=99$$
$$3a_1+30C=99$$
$$a_1+10C=33$$
$$a_3+a_{19}=\left(a_1+2C\right)+\left(a_1+18C\right)=33$$
$$=2a_1+20C$$
$$=2\left(a_1+10C\right)$$
Note that $$\left(a_1+10C\right)=33$$
$$2\left(33\right)$$
$$a_3+a_{19}=66$$
Answer is Option A
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sequence

by GMATGuruNY » Sun Jun 03, 2018 2:32 am
Gmat_mission wrote:In the sequence a_1, a_2, . . . , a_n each term after the first term is equal to the preceding term plus a constant c. $$a_1+a_{11}+a_{21}=99.$$ What is the value of $$a_3+a_{19}=?$$ A. 66
B. 44
C. 33
D. 22
E. 11
For any evenly spaced set:
average = median.
sum = (count)(average).

a� + a�� + a₂� = 99.
In the given sequence, the difference between one term and the next = c.
Implication:
{a�, a��, a₂�} constitutes an evenly spaced set because the difference between one term and the next = 10c.
Thus, the median of the set -- in other words, the value of a�� -- is equal to the average of the 3 terms:
a�� = sum/count = 99/3 = 33.

a�� is also the median -- and thus the average -- of {a₃, a�₉}.
Thus, the sum of these 2 terms = (count)(average) = (2)(33) = 66.

The correct answer is A.
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by Jeff@TargetTestPrep » Sun Jun 03, 2018 5:34 pm
Gmat_mission wrote:In the sequence a_1, a_2, . . . , a_n each term after the first term is equal to the preceding term plus a constant c. $$a_1+a_{11}+a_{21}=99.$$ What is the value of $$a_3+a_{19}=?$$ A. 66
B. 44
C. 33
D. 22
E. 11
a(11) = a(1) + 10c

a(21) = a(1) + 20c

Thus:

a(1) + a(1) + 10c + a(1) + 20c = 99

3a(1) + 30c = 99

a(1) + 10c = 33

Next, let’s evaluate a(3) + a(19).

a(3) + a(19) = a(1) + 2c + a(1) + 18c = 2a(1) + 20c = 2[a(1) + 10c] = 2 x 33 = 66.

Answer: A

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