sum of an arithmetic progression

**stressedIBstudent1** · December 6th 2010, 02:16 PM

The ratio of the 5th term to the 12th term of a sequence in an arithmetic progression is 6/13. If each of this sequence is positive, and the product of the 1rst term and the 3rd term is 32, find the sum of the first 100 terms of this sequence.

Hope someone can help, I'm studying for my math final.

Which formula should I use?

**e^(i*pi)** · December 6th 2010, 02:23 PM

You are told that $\dfrac{U_5}{U_{12}} = \dfrac{6}{13}$ and $U_1U_3 = 32$ .

By definition $a = U_1$ and $U_n = a+(n-1)d$

Next work out $U_5$ and $U_{12}$ and $U_3$ in terms of $a$ and $d$ and the solve the simultaneous equations

Edit: Sum of an arithmetic sequence is $S_n = \dfrac{n}{2}\left(2a+(n-1)d\right)$

**pickslides** · December 6th 2010, 02:24 PM

Before finding $\displaystyle S_{100}$ you need to find $\displaystyle a$ and $\displaystyle d$ such that

$\displaystyle\frac{t_5}{t_{13}}=\frac{6}{13}=\frac {a+(5-1)d}{a+(13-1)d}$

and

$\displaystyle t_1\times t_3 = t_3 \times a = 32$

Does this help?

**stressedIBstudent1** · December 6th 2010, 02:48 PM

Hey, thank you so much for helping me but I'm still not getting it, could you please explain further?

**e^(i*pi)** · December 6th 2010, 02:55 PM

Originally Posted by stressedIBstudent1

Hey, thank you so much for helping me but I'm still not getting it, could you please explain further?

You should know that $U_5 = a+4d$ and $U_{12} = a + 11d$

$\dfrac{a+4d}{a+11d} = \dfrac{6}{13} \text{.......... eq(1) }$

$U_1U_3 = a(a+2d) = 32 \text{.......... eq(2) }$

Solve simultaenously for a and d.

**Archie Meade** · December 6th 2010, 03:38 PM

Originally Posted by stressedIBstudent1

The ratio of the 5th term to the 12th term of a sequence in an arithmetic progression is 6/13. If each of this sequence is positive, and the product of the 1rst term and the 3rd term is 32, find the sum of the first 100 terms of this sequence.

Hope someone can help, I'm studying for my math final.

Which formula should I use?

$T_1=a,\;\;T_2=a+d,\;\;T_3=a+2d,\;\;T_4=a+3d,\;\;T_ 5=a+4d,\;\;.....T_{12}=a+11d$

$\displaystyle\frac{T_5}{T_{12}}=\frac{a+4d}{a+11d} =\frac{6}{13}\Rightarrow\ 13(a+4d)=6(a+11d)$

$\Rightarrow\ 13a+52d=6a+66d\Rightarrow\ 13a-6a=66d-52d\Rightarrow\ 7a=14d\Rightarrow\ a=2d$

Substitute this "a" into $a(a+2d)=32$

Alternatively work the other way...

$32=1(32)=2(16)=4(8)$ so you could test $a=2,\;\; d=7$ and $a=4,\;\;d=2$ in the ratio equation.

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