Question

Sum the following series to n terms 7+7.7+7.77+7.777+.........

Answer 1

Let me start with the second term.

7(1.1+1.11+1.111+...)

Multiply and divide by 9

=7/9(9.9+9.99+9.999+...)=79(9.9+9.99+9.999+..)

=7/9(10−0.1+10−0.01+10−0.001+...)
=7/9(10−0.1+10−0.01+10−0.001+...)

If there are n terms, then then the sum of the positive terms is 10n, and the negative terms form a geometric progression, with first term 0.1 and ratio 0.1. Using the formula for sum of geometric progression to n terms, we get

=7/9(10n−0.1(1−0.1n)/0.9)

Answer 2

Answer:

$\frac{7}{9} [(10n - \frac{(1-0.2^{n})}{1-0.1} ]$

The above equation is the sum of the series

$7+7.7+7.77+7.777+..............+nth$

Step-by-step explanation:

Very firstly, consider the given series up to n terms

Which is,

$7+7.7+7.77+7.777+..............+nth$ term,

Now, Take out 7 as common from the above series

We get,

$7(1+1.1+1.11+1.111+..............+nth)$     ..............(1)

Now, Multiply and Divide 9 in equation (1)

$\frac{7}{9} (9+9.9+9.99+9.999+............+nth)$

Let's simplify the above equation,

we get,

$\frac{7}{9} [(10-1)+(10-0.1)+(10.0.01)+(10-0.001)+............+nth)]$

Again,

$\frac{7}{9} [(10+10+10+10............+nth) -(1+0.1+0.01+0.001+........+nth)]$

Hence,

$\frac{7}{9} [(10n -(1+0.1+0.01+0.001+........+nth)]$  .........(2)

As we know that,

The Sum of a series in G.P. is:

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

where a is the first number of the series, and r is the common ratio of the series.

Applying on equation (2), we get

$\frac{7}{9} [(10n - \frac{(1-0.2^{n})}{1-0.1} ]$

The above equation is the sum of the series

$7+7.7+7.77+7.777+..............+nth$.

To know about G.P.  and A.P. series,

https://brainly.in/question/13385188

To know about the formula of G.P. series

https://brainly.in/question/10044188