# How do you find the sum of the infinite geometric series a1=26 and r=1/2?

By signing up, you agree to our Terms of Service and Privacy Policy

To find the sum of the infinite geometric series with first term ( a_1 = 26 ) and common ratio ( r = \frac{1}{2} ):

- Use the formula for the sum of an infinite geometric series: ( S = \frac{a_1}{1 - r} ).
- Substitute the given values: ( S = \frac{26}{1 - \frac{1}{2}} ).
- Simplify: ( S = \frac{26}{1 - \frac{1}{2}} = \frac{26}{\frac{1}{2}} = 26 \times 2 = 52 ).

Therefore, the sum of the infinite geometric series is 52.

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- Given that #f(x)=x^3+4x^2+bx+c#. When *f* is divided by #(x-3)#, the remainder is 110. In addition, when *f* is divided by #(x+2)#, the remainder is 150. The sum of #b+c=?#
- If #a_1=5# and #a_i = a_(i-1)+2#, what is the sum of the first 700 terms in the sequence?
- How do you find the sum of the arithmetic sequence given a1=3, d=-4, n=8?
- How do you find the sum of the series #i^2# from i=1 to 12?
- Summation of series(method of differences),When we use #∑f(r)-f(r+1#),how can we know it is #f(1)-f(n+1)#, but not #f(n)-f(1+1)#? Also, how can we know it is #f(n+1)-f(1)# but not #f(1+1)-f(n)# when we use #∑f(r+1)-f(r)#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7