Another AP/GP question

**margaritas** · September 3rd 2006, 01:38 AM

The sum of the first 100 terms of an arithmetic progression is 10000; the first, second and fifth terms of this progression are three consecutive terms of a geometric progression. Find the first term a and the non-zero common difference, d, of the arithmetic progression. (Answer: a = 1, d = 2)

I found:

2a + 99d = 200

After which, I'm stuck.

Thanks in advance if you could help with this question!

**Glaysher** · September 3rd 2006, 02:01 AM

Sum of first 100 terms = $\frac{100}{2}(2a+(100-1)d)=10000$

gives $2a + 99d = 200$

1st term = $a$

2nd term = $a+d$

5th term= $a+4d$

Three terms of a geometric so exists $r$ such that

$ar = a + d$ and $ar^2=a+4d$

So $ar^2 = ar + 3d$ by subing in first equation into second

Also $ar^2 = ar + rd$ by multiplying both sides of first equation by $r$

So $r=3$

So $3a = a+d$ and $2a = d$

Sub into equation you got $d + 99d = 100d = 200$

$d=2$

$a = 1$

**margaritas** · September 3rd 2006, 02:34 AM

Yay thanks glaysher now I get it!

**Glaysher** · September 3rd 2006, 02:46 AM

Originally Posted by margaritas

Yay thanks glaysher now I get it!

No problem

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