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Mathematics

Find the sum of first n terms of :

(i) 4 + 44 + 444 + ……

(ii) 0.7 + 0.77 + 0.777 + …….

AP GP

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Answer

(i) Given,

⇒ 4 + 44 + 444 + …… upto n terms

⇒ 4(1 + 11 + 111 + ……. upto n terms)

⇒ 49(9+99+999+......\dfrac{4}{9}(9 + 99 + 999 + …… upto n terms)

⇒ 49[(10−1)+(102−1)+......\dfrac{4}{9}[(10 - 1) + (10^2 - 1) + …… upto n terms]

⇒ 49[(10+102.......+10n)−(1+1+........n times)]\dfrac{4}{9}\Big[(10 + 10^2 ……. + 10^n) - (1 + 1 + …….. \text{n times})\Big]

Above sequence : 10 + 102 + ……… + 10n

It is a G.P. with common term (a) = 10 and common ratio (r) = 10.

By formula,

Sum of G.P. = a(1−rn)1−r\dfrac{a(1 - r^n)}{1 - r}

⇒ 49[10(10n−1)10−1−n]\dfrac{4}{9}\Big[\dfrac{10(10^n - 1)}{10 - 1} - n\Big]

⇒ 49[10(10n−1)9−n]\dfrac{4}{9}\Big[\dfrac{10(10^n - 1)}{9} - n\Big].

Hence, sum of 4 + 44 + 444 + …… upto n terms = 49[10(10n−1)9−n].\dfrac{4}{9}\Big[\dfrac{10(10^n - 1)}{9} - n\Big].

(ii) Given,

⇒0.7+0.77+0.777+....... upto n terms ⇒7(0.1+0.11+0.111+...... upto n terms )⇒79(0.9+0.99+0.999+....... upto n terms )⇒79[(1−0.1)+(1−0.01)+(1−0.001)+.......... upto n terms]⇒79[1+1+...... upto n terms −(0.1+0.01+0.001+........)]⇒79[n−(110+1100+11000+...... upto n terms)]\Rightarrow 0.7 + 0.77 + 0.777 + ……. \text{ upto n terms } \\[1em] \Rightarrow 7(0.1 + 0.11 + 0.111 + …… \text{ upto n terms }) \\[1em] \Rightarrow \dfrac{7}{9}(0.9 + 0.99 + 0.999 + ……. \text{ upto n terms }) \\[1em] \Rightarrow \dfrac{7}{9}[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + ………. \text{ upto n terms}] \\[1em] \Rightarrow \dfrac{7}{9}[1 + 1 + …… \text{ upto n terms } - (0.1 + 0.01+ 0.001 + ……..)] \\[1em] \Rightarrow \dfrac{7}{9} \Big[n - \Big(\dfrac{1}{10} + \dfrac{1}{100} + \dfrac{1}{1000} + …… \text{ upto n terms}\Big)\Big] \\[1em]

Above sequence : 110+1100+11000+.........\dfrac{1}{10} + \dfrac{1}{100} + \dfrac{1}{1000} + ………

It is a G.P. with first term (a) = 110\dfrac{1}{10} and common ratio (r) = 110\dfrac{1}{10}.

By formula,

Sum of G.P. = a(1−rn)1−r\dfrac{a(1 - r^n)}{1 - r}

⇒79[n−110(1−(110)n)1−110]⇒79[n−110(1−110n)910]⇒79[n−(1−110n)9]⇒79[n−19(1−110n)]\Rightarrow \dfrac{7}{9} \Big[n - \dfrac{\dfrac{1}{10}\Big(1 - \Big(\dfrac{1}{10}\Big)^n\Big)}{1 - \dfrac{1}{10}}\Big] \\[1em] \Rightarrow \dfrac{7}{9} \Big[n -\dfrac{\dfrac{1}{10}\Big(1 - \dfrac{1}{10^n}\Big)}{\dfrac{9}{10}}\Big] \\[1em] \Rightarrow \dfrac{7}{9} \Big[n -\dfrac{\Big(1 - \dfrac{1}{10^n}\Big)}{9}\Big] \\[1em] \Rightarrow \dfrac{7}{9} \Big[n - \dfrac{1}{9}\Big(1 - \dfrac{1}{10^n}\Big)\Big]

Hence, sum of 0.7 + 0.77 + 0.777 + ……. upto n terms = 79[n−19(1−110n)]\dfrac{7}{9} \Big[n - \dfrac{1}{9}\Big(1 - \dfrac{1}{10^n}\Big)\Big].

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