Name: Arithmetic Mean
Uploaded: 2019-07-19
Duration: 3 min 49 s
Description: Given two numbers, a and b, we can insert a number A between them so that a, A, b is an A.P. Such a number A is called the arithmetic mean (A.M.) of a and b....

Question

Find the sum to $n$ terms $8 + 88 + 888 + 8888 + \dots$

Accepted Answer

Let the given series $S_{n} = 8 + 88 + 888 + 8888 + \dots$ to $n$ terms

By observing that, the given series is not G.P.

Therefore making the series to G.P by as follows

$\Rightarrow S_{n} = 8 (1 + 11 + 111 + 1111 + . . .$ to $n$ terms) (Taking out common $8$ )

$\Rightarrow S_{n} = \frac{8}{9} (9 + 99 + 999 + 9999 + . . .$ to $n$ terms) (Multiplied and divided by $9$ )

$\Rightarrow S_{n} = \frac{8}{9} [(10 - 1) + (100 - 1) + (1000 - 1) + (10000 - 1) + . . .$ to $n$ terms] (Make $9 \Rightarrow (10 - 1), 99 \Rightarrow (100 - 1) . . . .$ so on ]

$\Rightarrow S_{n} = \frac{8}{9} [(10 + 10^{2} + 10^{3} + 10^{4} + \dots$ to $n$ terms) $- (1 + 1 + 1 + 1 + \dots n$ terms) $] \to (1)$

Here the first series is in G.P with $a = 10, r = \frac{100}{10} = 10$

We know that, the sum to $n$ terms in G.P is $S_{n} = \frac{a (r^{n} - 1)}{r - 1}, r > 1$

Hence, $(1)$ becomes $S_{n} = \frac{8}{9} [\frac{10 (10^{n} - 1)}{10 - 1} - n]$

$\Rightarrow S_{n} = \frac{8}{9} [\frac{10}{9} (10^{n} - 1) - n]$

$\Rightarrow S_{n} = \frac{8}{81} [10 (10^{n} - 1) - 9 n]$

Hence, the sum to $n$ terms of given series is $\frac{8}{81} [10 (10^{n} - 1) - 9 n]$ .

Find the sum to $n$ terms $8 + 88 + 888 + 8888 + \dots$

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

Find the sum to n terms 8+88+888+8888+…

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

Find the sum to $n$ terms $8 + 88 + 888 + 8888 + \dots$