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Exercise 5.2 · Q3

Q.Find the sum to the indicated number of terms in each of the geometric progressions:

(i) 3,3,33,…\sqrt{3}, 3, 3\sqrt{3}, \ldots 6 terms
(ii) 0.15+0.015+0.0015+…0.15 + 0.015 + 0.0015 + \ldots 20 terms.
Lakshadweep CbseNCERTSubjective· 3mImportance★★★★★est
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Using Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1}: the first G.P. sums to 39+13339+13\sqrt3 over 6 terms; the second sums to 16(1−10−20)\dfrac16(1-10^{-20}) over 20 terms.

Sum of nn terms of a G.P. (for r≠1r\ne1): Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1}, where aa is the first term and rr the common ratio.

(i) 3,3,33,…\sqrt3,3,3\sqrt3,\ldots; 6 terms

  1. a=3a=\sqrt3, r=33=3r=\dfrac{3}{\sqrt3}=\sqrt3.
  2. (3)6=33=27(\sqrt3)^6=3^3=27.
  3. S6=3(27−1)3−1=2633−1S_6=\dfrac{\sqrt3(27-1)}{\sqrt3-1}=\dfrac{26\sqrt3}{\sqrt3-1}.
  4. Rationalise: multiply by 3+13+1\dfrac{\sqrt3+1}{\sqrt3+1}: S6=263(3+1)3−1=26(3+3)2=13(3+3)=39+133S_6=\dfrac{26\sqrt3(\sqrt3+1)}{3-1}=\dfrac{26(3+\sqrt3)}{2}=13(3+\sqrt3)=39+13\sqrt3.
  5. Numerically: 39+13(1.732)≈39+22.52=61.5239+13(1.732)\approx39+22.52=61.52.
  6. Self-check by direct addition: 3+3+33+9+93+27≈1.73+3+5.20+9+15.59+27=61.52\sqrt3+3+3\sqrt3+9+9\sqrt3+27\approx1.73+3+5.20+9+15.59+27=61.52. ✓

(ii) 0.15+0.015+0.0015+…0.15+0.015+0.0015+\ldots; 20 terms

7. a=0.15a=0.15, r=0.0150.15=0.1r=\dfrac{0.015}{0.15}=0.1.

8. S20=0.15(1−(0.1)20)1−0.1=0.150.9(1−10−20)=16(1−10−20)S_{20}=\dfrac{0.15\big(1-(0.1)^{20}\big)}{1-0.1}=\dfrac{0.15}{0.9}\big(1-10^{-20}\big)=\dfrac16\big(1-10^{-20}\big).

9. Since 10−2010^{-20} is utterly negligible, S20≈16=0.16‾S_{20}\approx\dfrac16=0.1\overline{6}.

10. Self-check: for an infinite G.P. with ∣r∣<1|r|<1, S∞=a1−r=0.150.9=16S_\infty=\dfrac{a}{1-r}=\dfrac{0.15}{0.9}=\dfrac16; the 20-term sum is this same value less an utterly negligible correction, as expected. ✓

✓Final answer

(i) S6=39+133≈61.52S_6=39+13\sqrt3\approx61.52 (ii) S20=16(1−10−20)≈16S_{20}=\dfrac16(1-10^{-20})\approx\dfrac16

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