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Bunuel
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

\(t_{10}\) 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

\((n/2) * [2a + (n - 1)*d] = (9/2)*[116 + 8*6] = 9*82 = 738\)

Answer (C)
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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!
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Bunuel
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

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

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

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


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.
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Bunuel
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

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
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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.
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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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