If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =

Question

If S is the infinite sequence such that  t_1 = 4 ,  t_2 = 10 , …,  t_n = t_{n-1} + 6 ,…, what is the sum of all the terms from  t_{10}  to  t_{18} ?

(A) 671
(B) 711
(C) 738
(D) 826
(E) 991

DavidTutorexamPAL · Accepted Answer

As there are specific rules for straightforward calculation of sums of arithmetic sequences, we'll use them.
This is a Precise approach.

To calculate the sum of an arithmetic sequence, we need to know the first number, the last number and the number of elements in the sequence.
Our first number is t_10 which is equal to t_1 + d*9 = 4 + 6*9 = 58.
Our last number is t_18 which is equal to t_10 + d*8 = 58 + 6*8 = 106
As there are 9 total elements, our sum is (58+106)*9/2 = 164*9/2 = 82*9 = 82(10 - 1) = 820-82 = 738

(C) is our answer.

Note that as all elements of our sequence are even, we could also have eliminated (A), (B), (E) without calculation and then guessed bewteen (C) and (D).

Bunuel · Answer

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KarishmaB · Answer

t_{10}  to  t_{18}  can be considered an AP with first term as  t_{10} , common difference as 6 and with total 9 terms.

The first term,  t_{10} = t_1 + 9*6 = 58

Sum of 9 terms of AP with first term as 60 is

(n/2) *   = (9/2)*  = 9*82 = 738

Answer (C)

jfranciscocuencag · Answer

Hello VeritasKarishma

A basic question that is making circles in my head...

In order to calculate n (quantity), is it always (Last - First) + 1?

Ex. Number or terms between 10 and 18:

(18 - 10 ) + 1 = 9?

Kind regards!

NinetyFour · Answer

t_{10} = 4 + (10-1)*6 = 58

t_{18} = 4 + (18-1)*6 = 106

\frac{9}{2}* (2*58) + (9-1)*6 =738

KanishkM · Answer

jfranciscocuencag , Would like to pitch in

Your reasoning is correct.

from 5 to 10 = 10-5-1 = 4
6,7,8,9

between 5 and 10 = 10-5+1 = 6
5,6,7,8,9,10

You can always take small numbers to understand the logic.

ScottTargetTestPrep · Answer

We see we have an evenly spaced set with 6 being the difference between terms.

Since t1 = 6 x 1 - 2 = 4 and t2 = 6 x 2 - 2 = 10, we see that:

t10 = 6 x 10 - 2 = 58 and t18 = 6 x 18 - 2 = 106

Thus, the sum of the terms from t10 to t18 is:

(58 + 106)/2 x (18 - 10 + 1)

82 x 9 = 738

Alternate Solution:

We see that this is an arithmetic sequence, with common difference d = 6

The formula for the nth term of an arithmetic sequence is t_n = t_1 + (n – 1)d

Thus, t_10 = t_1 + (10 -1)6 = 4 + (9)(6) = 58

Since each succeeding term is 6 more than the previous, we have:

t_11 = 64, t_12 = 70, t_13 = 76, t_14 = 82, t_15 = 88, t_16 = 94, t_17 = 100 and t_18 = 106.

The sum of the terms from t_10 to t_18 is:

58 + 64 + 70 + 76 + 82 + 88 + 94 + 100 + 106 = 738

Answer: C

Parbhar95 · Answer

using the sequence formula, 
an=a1+(n-1)d, tn=4+(n-1)6

t10=4+(10-1)6=58
t18=4+(18-1)6=106

using average formula - avg=sum/# of terms

sum=avg x # of terms

because you have an evenly spaced set, avg=(1st term in sequence+last term in sequence)/2 
avg=(t10+t18)/2 = (58+106)/2 = 82.

82x9=738 so C.

harshittahuja · Answer

If S is the infinite sequence such that t1=4, t2=10, …, tn=tn−1+6,…, what is the sum of all the terms from t10to t18?
Solution:-
Here it is Ap with 
T1=4 
T10=4+6*9=58
T18=4+17*6=106
Sum =9/2(106+58).
=9/2(164)
=9*82
=738

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If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =

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