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Exercise 5.2 · Q1

Q.Find the indicated terms in each of the Geometric Progressions given below:

(i) 4,12,36,…4, 12, 36, \ldots; 5th term
(ii) 3,−1,13,−19,…3, -1, \dfrac{1}{3}, -\dfrac{1}{9}, \ldots; 4th term, nnth term
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Both parts use an=arn−1a_n=ar^{n-1}: the 5th term of the first G.P. is 324324; the second G.P.'s 4th term is −19-\dfrac19 with general term an=3(−13)n−1a_n=3\left(-\dfrac13\right)^{n-1}.

nnth term of a G.P.: an=arn−1a_n=ar^{n-1}, where aa is the first term and r=ak+1akr=\dfrac{a_{k+1}}{a_k} is the common ratio.

(i) 4,12,36,…4,12,36,\ldots; 5th term

  1. a=4a=4, r=124=3r=\dfrac{12}{4}=3.
  2. a5=ar5−1=4×34=4×81=324a_5=ar^{5-1}=4\times3^4=4\times81=324.
  3. Self-check: 4,12,36,108,3244,12,36,108,324 — the 5th listed term is indeed 324324. ✓

(ii) 3,−1,13,−19,…3,-1,\dfrac13,-\dfrac19,\ldots; 4th term and nnth term

4. a=3a=3, r=−13=−13r=\dfrac{-1}{3}=-\dfrac13 (check: 1/3−1=−13\dfrac{1/3}{-1}=-\dfrac13 ✓, −1/91/3=−13\dfrac{-1/9}{1/3}=-\dfrac13 ✓).

5. 4th term: a4=ar3=3(−13)3=3×(−127)=−19a_4=ar^3=3\left(-\dfrac13\right)^3=3\times\left(-\dfrac1{27}\right)=-\dfrac1{9}.

6. nnth term: an=arn−1=3(−13)n−1a_n=ar^{n-1}=3\left(-\dfrac13\right)^{n-1}.

7. Self-check: the listed 4th term is −19-\dfrac19, matching step 5. ✓

✓Final answer

(i) 324324 (ii) a4=−19a_4=-\dfrac19,  an=3(−13)n−1\ a_n=3\left(-\dfrac13\right)^{n-1}

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