An investment of $2500 is made in an account that compounds interest quarterly. After years 30 the balance in the account is $26132.78. To the nearest tenth of a percent, what is the annual interest rate for this account?

A) $2.0\%$
B) $0.5\%$
C) $31.5\%$
D) $7.9\%$
E) $2.6\%$

We can use the formula for compound interest:

$A=P\left(1+\frac{r}{n}\right)^{nt}$, where
- A is the ending balance
- P is the initial investment
- r is the annual interest rate (as a decimal)
- n is the number of times interest is compounded per year
- t is the number of years

Since interest is compounded quarterly, we have n=4 and t=30. Also, we are given that P=2500 and A=26132.78. So we can solve for r:

$26132.78 = 2500\left(1+\frac{r}{4}\right)^{4\cdot 30}$
$\frac{26132.78}{2500} = \left(1+\frac{r}{4}\right)^{120}$
$10.453112 = \left(1+\frac{r}{4}\right)^{120}$
$\ln 10.453112 = \ln\left(1+\frac{r}{4}\right)^{120}$
$\ln 10.453112 = 120\ln\left(1+\frac{r}{4}\right)$
$\frac{\ln 10.453112}{120} = \ln\left(1+\frac{r}{4}\right)$
$0.007784209 = \ln\left(1+\frac{r}{4}\right)$
$e^{0.007784209} = 1+\frac{r}{4}$
$1.007816156 = 1+\frac{r}{4}$
$\frac{r}{4} = 0.007816156$
$r = 0.031264624$ (as a decimal)

Finally, we convert the annual interest rate to a percentage and round to the nearest tenth:

$r \approx 3.1264624\% \approx 3.1\%$

So the answer is D, 7.9%.