Find the sum to n terms of the sequence, 8, 88, 888, 8888…

Asked by Pragya Singh | 1 year ago |  108

##### Solution :-

Given sequence: 8, 88, 888, 8888…

This sequence is not a G.P.

But, it can be changed to G.P. by writing the terms as

Sn = 8 + 88 + 888 + 8888 + …………….. to n terms

$$\dfrac{8}{9}[9+99+999+9999$$

$$+....to\;n\;terms]$$

$$\dfrac{8}{9}[(10-1)+(10^2-1)+(10^3-1)$$

$$+(10^4-1)+.......to\;n\;terms]$$

$$\dfrac{8}{9}[10+10^2+....to\;n\;terms$$

$$-(1+1+1+..n\;tems)]$$

$$\dfrac{8}{9}[\dfrac{10(10^n-1)}{10-1}-n]$$

$$\dfrac{8}{9}[\dfrac{10(10^n-1)}{9}-n]$$

$$\dfrac{80}{81}(10^n-1)-\dfrac{8}{9}n$$

Answered by Abhisek | 1 year ago

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