The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)

Question

The  n_{th}  term,  t_n , of a certain sequence is defined as  t_n = t_{n-1} + 4 . If  t_1 = -11 , then  t_{82} =

(A) 313
(B) 317
(C) 320
(D) 321
(E) 340

kunalcvrce · Answer

t2=t1+4 
t3=t2+4
.
.
t82=t81+4

adding all we get t82=t1+81*4, since t1=-11
t82=313.

option A

generis · Answer

This problem  looks  like a regular arithmetic progression.

The standard equation for AP is usually detectable in questions like this one.

t_n = t_{n-1} + 4

t_1 = -11 
 t_2 =(t_{1} + 4)=(-11+4)= -7 
 t_3 = t_{2} + 4 = -3 
 t_4 = t_{3} + 4 = 1

The above is equivalent to . . . (looking for standard arithmetic sequence)

t_1 = -11 
 t_2 = t_1 + 4 = -7 
 t_3 = t_1 + 4 + 4 = -3 
 t_4 = t_1 + 4 + 4 + 4 = -1

Extrapolate the standard equation of arithmetic sequence. 
The number of 4s added is (n - 1)
4 = common difference, d

a_n = a_1 + (n-1)d

Rewrite:
 t_n = t_1 + (n-1)4 
 t_1  is -11

Hence
 t_{82} = -11 + (81*4) = 
 (-11 + 324) = 313

Answer A

mansi1997 · Answer

Can someone explain how did we calculate the number of terms here (i.e 81)?

SiddharthR · Answer

Hello Mansi, Refer to this section : https://gmatclub.com/forum/sequences-ma ... 62490.html What this question basically is --> t(1)+t(2)+t(3)+ ------ + t(82) -->(i) Given t1 = -11 and t(n) = t(n-1) + 4 (Used to calculate any term in the sequence) So when we used the t(n) eqn and t(1) = - 11 with (i), we get a sequence such as : (-11) + (-7) + (-3) + 1 + 5+ 9 + ---- ; Each term here is 4 greater than the previous term (noticed in the t(n) eqn above); This is an Arithmetic Progression (Constant distance between two successive terms). To find the 82nd value, t(82) = t(82 - 1) + 4 = t(81) + 4, you assume we may need the 81st value; In a way that is correct which means that you need to calculate each and every term to reach the 82nd value which is tedious. Instead, there is an easier approach to this problem. Each of the terms as mentioned above is separated from the other term by a constant distance of 4; This implies that you can write each of the successive terms in the sequence in terms of the first like the following --> t1, t2 = t1+4, t3 = t1+2(4), t4 = t1+3(4) + -----; As you move from one term to the next, the value increases by a multiple "4". Therefore with the 82nd term, we can write t(82) = t1 + 81(d) = -11 + 81(4) = 313 The above equation is generalized in the link above. Have a go through it. It's a great read.

ScottTargetTestPrep · Answer

Solution:

From t(n) = t(n - 1) + 4, we see that each term, starting from the 2nd term, is 4 more than the previous term. Since t(1) = -11, we have:

t(2) = -7, t(3) = -3, t(4) = 1, and so on.

Note that this is an arithmetic sequence with first term t(1) = -11 and common difference d = 4. Using the formula for the nth term of an arithmetic sequence, we have t(n) = t(1) + d(n - 1). Substituting, we have:

t(n) = -11 + 4(n - 1) for n ≥ 2

So, t(82) = -11 + 4(82 - 1) = -11 + 324 = 313.

Answer: A

Babineaux · Answer

One of the simplest way which I think is,

t(n) = t(n-1) + 4;

Thus, t(n) - t(n-1) = 4, which is nothing but the common difference, d.

Now, a, the first term is -11, given in the prob as t(1).

And t(n) = a + (n-1)  *  d

Putting values, t(82) = -11 + (82 - 1)  *  4,

Results in -11 + 81  *  4 = -11 + 324 = 313

Hence, ans is A.

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The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)

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