If the sum of the first five terms of an Arithmetic sequence is equal

Question

If the sum of the first five terms of an Arithmetic sequence is equal to 120 and the sum of the next five terms of the same Arithmetic Sequence is equal to 245, what is the 4th term of this Sequence?

A) 29
B) 34
C) 81
D) 86
E) 91

ArunSharma12 · Accepted Answer

sum of first five terms of an Arithmetic sequence =  x+(x+d)+(x+2d)+(x+3d)+(x+4d) =5x+10d
sum of next five terms of the Arithmetic sequence =  (x+5d)+(x+6d)+(x+7d)+(x+8d)+(x+9d) =5x+35d 
 5x+10d=120  --a
 5x+35d=245  --b
subtracting equations b and a,
25d = 125; d = 5

x+2d = 24 -- from equation a
4th term: x+3d = 24 + 5= 29
Ans: A

gracie · Answer

mean and 3rd term of first five terms=120/5=24
mean and 3rd term of next five terms=245/5=49
let d=common difference
24+d=49-4d (note that 49 is  8th  term of total sequence)
d=5
4th term of sequence=24+5=29
A

TestPrepUnlimited · Answer

Set the difference between each term as x.  Compared terms #6~10 to terms #1~5. #6 is 5x bigger than #1, and #7 is 5x bigger than #2 etc. Then the entire sum of #6~10 is is 25x bigger than the sum of #1~5. 
So  25x = 245 - 120 = 125  and  x = 5 .

Next, we can find the 3rd term quickly given the first 5 terms has a sum of 120.  The median and mean of those five terms are 120/5 = 24 which is equivalent to the 3rd term.  Then the 4th term is 24 + 5 = 29.

Ans: A

varun325 · Answer

If the sum of the first five terms of an Arithmetic sequence is equal to 120 and the sum of the next five terms of the same Arithmetic Sequence is equal to 245, what is the 4th term of this Sequence?

{2a+(n-1)d}n/2= Sum of first five terms
{2a+4d}5/2=120
2a+4d=48.............eq-1

{2a+(n-1)d}n/2=  Sum of first 10 terms 
{2a+9d}5=365 from(120+245)
2a+9d=73.........eq-2

Solving the two eq for d we get d=5 
put the value d=5 in eq-1 we get a=14
1st term=14
2nd term=19 ( as the difference in two terms of A.p is 5)
3rd term=24 
4th term=29 
ANS A

randomavoidplease · Answer

First up, This is my first 700+ question that I have gotten right. So I am super happy about that. 
Next the way I solved it,  Please Tell Me If This Actually Works or Was it just a fluke?  In my head it should but would love to know if there is something I missed.

Ok so list 1 = a+b+c+d+e=120
List 2 = f+g+h+J+k=245

First, 120 =  \frac{a+e}{2}  * 5 
Thus a+e = 48
Second 245 =  \frac{f+k}{2}  * 5 
Thus f+k = 98.
Difference = 98-48 = 50
Because this was a total of 10 number difference, I divided it by 10 to get each number difference, which is 5. 
The average of a+e = 48/2 = 24 + 5 = 29.

ScottTargetTestPrep · Answer

Solution:
 
We can let the first term = x and the common difference = d. Thus, the second term = x + d, the third term = x + 2d, and so on. We can create the equations:

x + (x + d) + (x + 2d) + (x + 3d) + (x + 4d) = 120 → 5x + 10d = 120

and
 
(x + 5d) + (x + 6d) + (x + 7d) + (x + 8d) + (x + 9d) = 245 → 5x + 35d = 245

Subtracting the first equation from the second equation, we have:

25d = 125

d = 5

Substituting d = 5 into the first equation, we have:

5x + 50 = 120

5x = 70

x = 14

Since the fourth term is x + 3d, the fourth term is 14 + 3(5) = 29.

Answer: A

Bambi2021 · Answer

One interesting observation.

The difference between the sums is equal to the 4th term times the number of terms in each set. Can anyone explain this fact in mathematical terms? Is this a coincidence?

I mean that 245-120 = 145 and 145/5=29.

ravigupta2912 · Answer

I used estimation to solve.

Combined sum of 10 terms is 365. Avg is 36.5.

An AP is equally spaced with median = mean. 36.5 is the avg of 5th and 6th terms with 6th higher than 5th.

So 5th term is lower than 36.5.

This rules C, D and E. between A and B, A is more likely since if 4th term is 32 something then the sun of first five terms is unlikely to be 120.

Admittedly not the best approach, but can come in handy.

Posted from my mobile device

CEdward · Answer

If the sum of the first five terms of an Arithmetic sequence is equal to 120 and the sum of the next five terms of the same Arithmetic Sequence is equal to 245, what is the 4th term of this Sequence?

A) 29
B) 34
C) 81
D) 86
E) 91

a1 + (a1 + d) + (a1 + 2d) + (a1 + 3d) + (a1 + 4d) = 120 --> 5a1 + 10d = 120

Next,

(a1 + 5d) + (a1 + 6d) + (a1 + 7d) + (a1 + 8d) + (a1 + 9d) = 245 --> 5a1 + 35d = 245

5a1 + 35d = 245
5a1 + 10d = 120 
---------------
25d = 125
d = 5

5a1 + 10(5) = 120
a1 = 14

Then,

14 + 3(5) = 29

A.

Crytiocanalyst · Answer

sum of first five terms = x+(x+d)+(x+2d)+(x+3d)+(x+4d)x+(x+d)+(x+2d)+(x+3d)+(x+4d)=5x+10d
next 5 terms = (x+5d)+(x+6d)+(x+7d)+(x+8d)+(x+9d)(x+5d)+(x+6d)+(x+7d)+(x+8d)+(x+9d)=5x+35d
5x+10d=1205x+10d=120 --1
5x+35d=2455x+35d=245 --2
Solving for A and B 
25d = 125; d = 5

x+2d = 24 from 1
4th term: x+3d = 24 + 5= 29
Hence IMO A

bumpbot · Answer

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If the sum of the first five terms of an Arithmetic sequence is equal

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