The infinite sequence a1, a2, a3, ... is defined by a1 = 3, a2

Question

The infinite sequence  a_1 ,  a_2 ,  a_3 , ... is defined by  a_1 = 3 ,  a_2 = 1 ,  a_3 = 2 , and  a_4 = -7 .

For each integer  n ≥ 5 :

if n is not divisible by 4, then  a_n = a_{(n−4)} + 3 
if n is divisible by 4, then  a_n = a_{(n−4)} − 8

What is the sum of the first 100 terms of the sequence?

A. 250
B. 275
C. 300
D. 325
E. 400

sujoykrdatta · Answer

For the first 4 terms, sum = -1

For the next 4 terms, we need to add 3 to the first 3 terms (since a5, a6, a7 i.e. 5,6 and 7 are not divisible by 4) and we need to subtract 8 once (from a8, since 8 is divisible by 4), thus adding a net value of 1. But the original sum of the terms is -1 and to that, we have added 1, giving the final sum 0.

For the next 4 terms, we need to add 3 to the first 3 terms (since a9, a10, a11 i.e. 9,10 and 11 are not divisible by 4) and we need to subtract 8 once (from a12, since 12 is divisible by 4), thus again adding a net value of 1. But the original sum of the terms is 0 (sum of the terms a5,a6,a7, a8) and to that, we have added 1, giving the final sum 1.

This will happen 24 times since after the first 4 terms, there are 24 more sets till 100 terms.

Thus, total 
= -1 + 0 + (1+2+... 23)
= -1 + 23×24/2 
= -1 + 276
= 275

Answer B

Dereno · Answer

The first four terms of the sequence is given:

a1 = 3

a2 = 1

a3 = 2

a4 = -7

The sum of these four values =  -1 .

Solving using the criteria given we get ,

a5 = 6

a6 = 4

a7 = 5

a8 = -15

The sum of these four values =  0

The next 4 values are (a9, a10, a11, a12) = (9, 8, 7, -23). The sum of these 4 values = (9+8+7-23) =  +1.

The next 4 values, just to check the sequence

(a 13, a14, a15, a16) = (12, 10, 11, -31). The sum of these 4 values = (12+10+11-31) =  +2.

So, for every 4 values, the sum gets incremented by +1.

(-1, 0, +1 , +2 , ....... Till what value?)

100 terms divided into blocks of 4, we get 25 blocks.

The initial two values are -1 and 0. So, we are left with 23 more.

(-1, 0, +1 , +2 , ....... +23)

From 1 to 23 , the sum is given by  n * (n+1)/2  = (23*24)/2 = 276

276-1+0 =  275 .

OR

(n/2) * ( First term + Last term )  = (25/2) * ( -1 + 23) = 25*11 =  275 .

OR

first term = a = -1

difference = d = 1.

Sum = (n/2) * (2a + (n-1)d)

= (25/2) * (2*-1 + 24*1)

= (25/2)*22

=  275 .

Option B .

pappal · Answer

in such questions we need to find a pattern when we start putting values after n>=5
the sequence is as: 3,1,2,-7 6,4,5,-15 9,7,8,-23.......
here summation of each of the set of four values is ONE more than the previous set starting from -1.
i.e. -1,0,1,2.....which goes upto 23
since 100-8(for the first two sets)=92/4=23 such sets after the first two.
hence summation of all the values=(23*24)/2 -1=276-1=275 
B

The infinite sequence a1, a2, a3, ... is defined by a1 = 3, a2

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The infinite sequence a1, a2, a3, ... is defined by a1 = 3, a2

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