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Exercise 5.3 · Q1

Q.Find the sum of the G.P.: 810,8100,81000,810000,…\dfrac{8}{10}, \dfrac{8}{100}, \dfrac{8}{1000}, \dfrac{8}{10000}, \ldots to nn terms.

(a) −89(110n−1)\dfrac{-8}{9}\left(\dfrac{1}{10^n}-1\right)
(b) 881(110n−1)\dfrac{8}{81}\left(\dfrac{1}{10^n}-1\right)
(c) 98(110n−1)\dfrac{9}{8}\left(\dfrac{1}{10^n}-1\right)
(d) 890(110n−1)\dfrac{8}{90}\left(\dfrac{1}{10^n}-1\right)
Yanam CbseNCERTSubjective· 1mImportance★★★★★est
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✓ Free question

Summing the GP 810,8100,…\dfrac{8}{10},\dfrac{8}{100},\ldots to nn terms gives −89(110n−1)-\dfrac{8}{9}\left(\dfrac{1}{10^n}-1\right) — option (a).

[!FORMULA] Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r}

aa = first term, rr = common ratio, nn = number of terms.

  1. Here a=810a=\dfrac{8}{10} and each term is 110\dfrac{1}{10} of the previous one, so r=110r=\dfrac{1}{10}.
  2. Substitute into the sum formula: Sn=810(1−110n)1−110=810(1−110n)910S_n=\dfrac{\frac{8}{10}\left(1-\frac{1}{10^n}\right)}{1-\frac{1}{10}}=\dfrac{\frac{8}{10}\left(1-\frac{1}{10^n}\right)}{\frac{9}{10}}.
  3. The 110\dfrac{1}{10} in numerator and denominator cancel: Sn=89(1−110n)S_n=\dfrac{8}{9}\left(1-\dfrac{1}{10^n}\right).
  4. Rewrite by flipping the sign inside: 89(1−110n)=−89(110n−1)\dfrac{8}{9}\left(1-\dfrac{1}{10^n}\right) = -\dfrac{8}{9}\left(\dfrac{1}{10^n}-1\right), matching option (a).
✓Final answer

(a) Sn=−89(110n−1)S_n=-\dfrac{8}{9}\left(\dfrac{1}{10^n}-1\right).

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