Arithmetic and geometric progression

[tongani.tony]

COURSE WORK PLEASE HELP ME OUT!!!!!!!


QN.1.....the sum of the series 1+8+15+...... is 396 .How many terms does the series contain?


QN.2 .....The sum of the first 2 terms of arithmetic progression is 40. The sum of the first 4 of the same arithmetic progression is 130.Determine the sum of the first 5 terms of the arithmetic progression?


QN.3...The sum of the first 2 terms of Geometric progression is 40. The sum of the first 4 terms of the same Geometric progression is 130. Find the 2 possible values of the sum of the first 5 terms of the geometric progression.

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[Guest]

1.....the sum of the series 1+8+15+...... is 396 . How many terms does the series contain?

Formula is
[tex]S= \frac{1}{2}(2a_1 + (n-1)d)n[/tex]

[tex]S = 396[/tex]
[tex]a_1 = 1[/tex]
[tex]d = 8 - 1 = 7[/tex]
We need to find n = ?

[tex]396 = \frac{1}{2}(2 + 7(n-1))n[/tex]

[tex]792 = 2n +7n(n-1)[/tex]
[tex]792 = 2n + 7n^2 - 7n[/tex]
[tex]792 = 7n^2 - 5n[/tex]
[tex]-7n^2 + 5n + 792 = 0[/tex]

[tex]7n^2 - 5n - 792 = 0[/tex]

We need to solve this equation there are 2 roots: 11 and [tex]-\frac{72}{7}[/tex]

Obviously the number of temrs must be a positive number so the number of terms is 11.

[Guest]

QN.2
[tex]\begin{array}{|l} a_{1 } + a_{2 } = 40 \\ a_{3 }+ a_{4 } = 90 \end{array}[/tex]

[tex]\begin{array}{|l} a_{1 } +a_{1 } +d= 40 \\ a_{1 }+2d+ a_{1 } +3d= 90 \end{array}[/tex]

[tex]\begin{array}{|l} 2 a_{1 } +d= 40 (A) \\ 2 a_{1 }+5d = 90 \end{array}[/tex]

4d=50 ; d= 12,5
See (A) 2[tex]a_{1 }[/tex]+12,5=40 ;[tex]a_{1 }[/tex]=13,75

[tex]S_{n }= \frac{2 a_{1 }+(n-1)d }{2} .n[/tex]

[tex]S_{5 }= \frac{2.13,75+4.12,5}{2} .5 =193,75[/tex]