Advertisements
Advertisements
प्रश्न
Find the sum of G.P. :
`1 - 1/3 + 1/3^2 - 1/3^3 + .........` to n terms.
योग
Advertisements
उत्तर
Given G.P. : `1 - 1/3 + 1/3^2 - 1/3^3 + .........` upto n terms
Here,
First term, a = 1
Common ratio, r = `(-1/3)/1 = -1/3` ...(∵ r < 1)
Number of terms to be added = n
∴ `S_n = (a(1 - r^n))/(1 - r)`
`=> S_n = (1(1 - (-1/3)^n))/(1 - (-1/3))`
= `(1(1 - (-1/3)^n))/(1 + 1/3)`
= `[1 - (-1/3)^n]/(4/3)`
= `3/4[1 - (-1/3)^n]`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Geometric Progression - Exercise 11 (D) [पृष्ठ १६१]
