Question

consider an infinite geometric series with first term a and common ratio r .If its sum is 4 and 2nd term is 3/4 then a=​

Answer 1

Answer:

a = 1, 3

Step-by-step explanation:

$\frac{a}{1 - r} = 4$

Cross-multiplying and writing in terms of r, we get:

$r = \frac{4 - a}{4}$

Now we know that second term is 3/4.

ar = 3/4

Substituting r in that we get:

$a( \frac{4 - a}{4} ) = \frac{3}{4}$

Solving the quadratic for a:

$a = 1$

$a = 3$

It can be either 1 or 3

Answer 2

Step-by-step explanation:

In an infinite GP, the sum of all terms is a/(1-r).

Also, the second term is ar=3/4.

By solving them,

$\frac{a}{1 - r} = 4 \\ a = 4 - 4r \\ ar = \frac{3}{4} \\ r(4 - 4r) = \frac{3}{4} \\ 4r - 4 {r}^{2} = \frac{3}{4} \\ 4 {r}^{2} - 4r + \frac{3}{4} = 0 \\ 16 {r}^{2} - 16r + 3 = 0 \\ r = \frac{24}{32} = \frac{3}{4} \\ \: or \: r = \frac{8}{32} = \frac{1}{4}$

So, r=3/4 or 1/4.

Case 1:

If r=3/4, then a=1.

Case 2:

If r=1/4, then a=3.

Thank you