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Exercise 5.2 · Q22

Q.The sum of an infinite G.P. is 3 and the sum of the squares of its terms is also 3, then its first term and common ratio are:

(i) 1,121, \dfrac{1}{2}
(ii) 12,32\dfrac{1}{2}, \dfrac{3}{2}
(iii) 32,12\dfrac{3}{2}, \dfrac{1}{2}
(iv) 1,141, \dfrac{1}{4}
Andaman Nicobar CbseNCERTSubjective· 1mImportance★★★★★est
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Combining the sum and sum-of-squares conditions for an infinite GP gives a=32, r=12a=\dfrac32,\ r=\dfrac12 — option (iii).

[!FORMULA] S∞=a1−rS_\infty=\dfrac{a}{1-r}; the squared terms a2, a2r2, a2r4,…a^2,\,a^2r^2,\,a^2r^4,\ldots also form a GP: ∑a2r2k=a21−r2\displaystyle\sum a^2r^{2k}=\dfrac{a^2}{1-r^2}

aa = first term, rr = common ratio (∣r∣<1|r|<1).

  1. Sum of the GP: a1−r=3  ⇒  a=3(1−r)\dfrac{a}{1-r}=3 \;\Rightarrow\; a=3(1-r). — (i)
  2. Sum of the squares of the terms: a21−r2=3\dfrac{a^2}{1-r^2}=3. — (ii)
  3. Substitute (i) into (ii): 9(1−r)2(1−r)(1+r)=3  ⇒  9(1−r)1+r=3\dfrac{9(1-r)^2}{(1-r)(1+r)}=3 \;\Rightarrow\; \dfrac{9(1-r)}{1+r}=3. …

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