Question

Find the sum of the eight terms of the series 1,2.5,6.25,…, coerrect to 1 decimal place.

Answer 1

Answer:

The required sum is

$=\frac{2}{3}((\frac{5}{2})^8-1)$

Step-by-step explanation:

Formula used:

Sum of the G.P a, ar, ar²..........is

$S_n=\frac{a(r^n-1)}{r-1}$

1, 2.5, 6.25..................

clearly it is an G.P with a=1, r=2.5

Sum of the eight terms

$=S_8$

$=\frac{a(r^n-1)}{r-1}$

$=\frac{1((2.5)^8-1)}{2.5-1}$

$=\frac{((\frac{5}{2})^8-1)}{1.5}$

$=\frac{((\frac{5}{2})^8-1)}{\frac{3}{2}}$

$=\frac{2}{3}((\frac{5}{2})^8-1)$

Answer 2

Answer:

the sum of the eight terms of the series 1,2.5,6.25,…, coerrect to 1 decimal place. = 1016.6

Step-by-step explanation:

Below are eight terms of the series 1,2.5,6.25,…,

1

2.5

6.25

15.625

39.0625

97.65625

244.140625

610.3515625

Adding all we get = 1016.5859375

up to one decimal = 1016.6