Question

In a GP the sum of the first two terms is 70 and the sum of the first six terms is 910. Find the sum of the first four terms

Answer 1

@jim_thompson5910 an easy one

Answer 2

what did you get?

Answer 3

i cant get it

Answer 4

I'm still thinking it through

Answer 5

i got 280 but i still have to figure it out.. I'm using shift solve though..

Answer 6

well if you use

\[\Large S_{n} = a*\frac{1-r^n}{1-r}\]

you can set up

\[\Large S_{2} = a*\frac{1-r^2}{1-r} = 70\]
\[\Large S_{6} = a*\frac{1-r^6}{1-r} = 910\]

Answer 7

ive done that but my anwer is wierd

Answer 8

Divide the two equations to get

\[\Large \frac{1-r^2}{1-r^6} = \frac{70}{910}\]

now solve for r

Answer 9

\[S_n = a_1\frac{ 1-r^n }{ 1-r }\]
\[70 = a_1\frac{ 1-r^2 }{ 1-r }\]
\[a_1 = \frac{ 70(1-r) }{ 1-r^2 }\]
\[910 = a_1\frac{ 1-r^6 }{ 1-r }\]
\[910 = \left( \frac{ 70(1-r) }{ 1-r^2 } \right)\left( \frac{ 1-r^6 }{ 1-r } \right)\]
\[\frac{ 910 }{ 70 } = \frac{ 1-r^6 }{ 1-r^2 }\]
\[13 = \frac{ 1-r^6 }{ 1-r^2 }\]
is there a way to get the value of 'r' without the help of a shift solve function in the calculator @jim_thompson5910 ?

Answer 10

it turns out that

\[\Large 1-r^6 = 1-(r^3)^2\]

\[\Large 1-r^6 = (1-r^3)(1+r^3)\]

\[\Large 1-r^6 = (1-r)(1+r+r^2)(1+r^3)\]

\[\Large 1-r^6 = (1-r)(1+r+r^2)(1+r)(1-r+r^2)\]

\[\Large 1-r^6 = (1-r)(1+r)(1+r+r^2)(1-r+r^2)\]

\[\Large 1-r^6 = (1-r)(1+r)(r^4+r^2+1)\]

so

\[\Large \frac{1-r^2}{1-r^6} = \frac{70}{910}\]

\[\Large \frac{(1-r)(1+r)}{(1-r)(1+r)(r^4+r^2+1)} = \frac{70}{910}\]

\[\Large \frac{1}{r^4+r^2+1} = \frac{70}{910}\]

\[\Large \frac{1}{r^4+r^2+1} = \frac{1}{13}\]

\[\Large r^4+r^2+1 = 13\]

This is a biquadratic which can be solved exactly

Answer 11

how do you solve it, never seen this b4

Answer 12

let z = r^2, so z^2 = r^4

you'll get

\[\Large z^2 + z+1 = 13\]

solve for z, then use the solutions for z to get the solutions for r

Answer 13

how don't get it isnt it no answer

Answer 14

what solutions for z did you get?

Answer 15

use the quadratic formula