Thank you for helping! I am soooooooo frustrated with arithmetic and geometric progression because I keep getting stuck at it. sigh.
Q1: The sum to n terms of a particular series is given by Sn = 17n - 3(n^2). Show that the series is an arithmetic progression and state the value of it's common difference.
Q2: The sum of the frist 3 terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121. Find the number of terms in this sequence.
Thank you!
A brutal approach:
"The sum of the frist 3 terms .... of an arithmetic progression are 9":
.... (1)
"The sum of ..... the last three terms of an arithmetic progression are ..... 57":
.... (2)
"It is known that the sum of the whole sequence is 121":
.... (3)
Substitute (1) into (2): .... (4)
Substitute (1) and (4) into (3) and solve for n.
Hello, Tangera!
These are challenging problems . . . Thank you!
The formula is: .Q1) The sum to terms of a particular series is given by:
Show that the series is an A.P. and find its common difference.
We must hammer into this form
We have: .
Add and subtract 6: .
And we have: .
. . Therefore: .
The AP is: .
I hope there is a shorter solution!Q2) The sum of the first 3 terms and that of the last three terms of an A.P.
are 9 and 57, respectively. It is known that the sum of the whole sequence is 121.
Find the number of terms in this sequence.
The first three terms are: .
Their sum is 9: .
The last three terms are: .
Their sum is 57: .
The entire sum is 121: .
Substitute [1] into [2]: .
Substitute [1] into [3]: .
Substitute [4] into [5]: ,
. . Therefore: .
The AP is: .
Too fast for me, Mr. F . . . Good job!