Question

The sum of three consecutive terms of a GP is 7.If each of the terms at the extremities is multiplied by 4 and the middle term is multiplied by 5 then the resulting numbers are in AP. Find the terms​

Answer 1

$Let \: the \: terms \: be \: a, \: ar, \:, {ar}^{2}$

$By \: question$

$a + ar + {ar}^{2} = 70$

$a(1 + r + {r}^{2} ) = 70$

$1 + r + {r}^{2} = \frac{70}{a} ....1)$

Again by question 4a,5ar,4ar² are in AP

$2(5ar) = 4a + 4a {r}^{2}$

$10ar = 4a(1 + {r}^{2} )$

$\frac{10ar}{4a} = 1 + {r}^{2}$

$\frac{5}{2} r = 1 + {r}^{2}$

Now, by adding r both sides,

$\frac{5}{2} r + r = 1 + r + {r}^{2}$

$\frac{7r}{2} = \frac{70}{a} (by \: 1)$

By cross multiplying,

$7r \times a = 70 \times 2$

$7ar = 140$

$a = \frac{140}{7r}$

$a = \frac{20}{r} .....2)$

Now by putting the value of a in 1)

$1 + r + {r}^{2} = \frac{70}{a}$

$1 + r + {r}^{2} = 70 \div \frac{20}{r}$

$1 + r + {r}^{2} = 70 \times \frac{r}{20}$

$20(1 + r + {r}^{2} ) = 70r$

$20 + 20r + 20 {r}^{2} = 70r$

$20 + 20 {r}^{2} = 50r$

$20 {r}^{2} - 50r + 20 = 0$

$10(2 {r}^{2} - 5r + 2) = 0$

$2 {r}^{2} - 5r + 2 = 0$

$2 {r}^{2} - (4 + 1)r + 2 = 0$

$2 {r}^{2} - 4 r- r + 2 = 0$

$2r(r - 2) - 1(r - 2) = 0$

$(r - 2)(2r - 1) = 0$

$r = 2 \: or \: r = \frac{1}{2}$

$if \: r = 2 \: then \: a = 10$

$and \: if \: r = \frac{1}{2} then \: a = 40$

Therefore the required terms are 10,20,40(when r=2 and a=10)

And,

The terms are 40,20,10(when r=1/2 and a=40)

Answer 2

Let the three consecutive terms be a,ar,ar²

given that, the sum of this three terms is 7

i.e a+ar+ar²=7...........(1)

also given that the extremities is multiplied by 4 & middle term is multiplied by 5

i.e 4a,5ar,4ar²

so, 2(5ar)=4a+4ar² [since, 2b=a+c]

➡2a(5r)=2a(2+2r²)

➡5r=2+2r²

➡2r²-5r+2=0

➡2r²-4r-r+2=0

➡(2r-1)(r-2)=0

so, r=1/2 or r=2

Now, (1)➡

when r=1/2

➡a+a(1/2)+a(1/2)²=7

➡a+(a/2)+(a/4)=7

➡7a=28 {taking LCM}

➡a=4

when,r=2

➡a+2a+4a=7

➡7a=7

➡a=1

Now the terms are[when a=1 & r=2]

1,2,4

and when a=4 & r=1/2

4,2,1

so the terms be 1,2,4