Sum of a number of terms in g.p., Mathematics

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We know that the terms in G.P. are:

a, ar, ar2, ar3, ar4, ................, arn-1

Let s be the sum of these terms, then

s = a + ar + ar2 + ar3 + ar4 + ................+ arn-1

                            or

s

 = 1058_GP.png

This is obtained as follows:

We know that

s = a + ar + ar2 + ar3 + ar4 +.............+ arn - 1                                            ......(1)

Multiplying this with "r" throughout, we have

r.s     = r.a + r.ar + r.ar2 + r.ar3 + r.ar4 +........+ r.arn-1 

         = ar + ar2 + ar3 + ar4 + ar5 +.......+ arn                                             ......(2)

Subtracting (1) from (2), we have

r.s - s = (ar - a) + (ar2 - ar) + (ar3 - ar2) +.......+ (arn - arn-1)

After canceling the terms equal in magnitude but opposite in sign, we are left with

s(r - 1) = arn - a
s(r - 1) = a(rn - 1)
 or s    = 427_GP1.png

By changing the signs in the numerator and the denominator we can also write the above equation as

 s  = 1058_GP.png

What happens to the above formula if the value of n is very large. The above formula can be written as

s

916_GP2.png

As the value of n approaches infinity (very large) the expression  2050_GP3.png becomes smaller to that extent where we ignore it. In this case the nth term is given as

Tn

= 1193_GP4.png

Now we look at a couple of examples.

Example

Find the sum of the series which is given below to 13 terms.

                   81, 54, 36, .............

The first term 'a' = 81 and the common ratio is obtained from the ratio of 54 and 81 or 36 and 54. It is 54/81 = 2/3. Now we employ the formula given above to calculate the sum of series to 13 terms.

s =   

1772_GP5.png

       = 241.78

The same series if considered as an infinite series, the sum of n terms would be

T = 491_GP6.png

= 243

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