Skip to content
Worked Examples · Example 19

Q.Find the sum of the indicated number of terms in the following G.P.

(i) 3,6,9,…3, 6, 9, \ldots 10 terms, nn terms
(ii) 1,−3,9,−27,…1, -3, 9, -27, \ldots 8 terms, pp terms
Yanam CbseNCERTSubjective· 2mImportance★★★★★est
64% · 68/106 Questions
✓ Free question

Apply Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1} to each G.P.; part (i) is read as 3,6,12,24,…3,6,12,24,\ldots (the ratio fixed by the first two terms) and part (ii) is 1,−3,9,−27,…1,-3,9,-27,\ldots.

[!FORMULA] Sum of first nn terms of a G.P. with first term aa and common ratio r≠1r\neq1:

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

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

Part (i): 3,6,12,24,…3,6,12,24,\ldots

  1. The ratio fixed by the first two terms is r=63=2r=\dfrac{6}{3}=2 (the standard textbook continuation 3,6,12,24,…3,6,12,24,\ldots is taken, since 3,6,93,6,9 has no constant ratio and cannot be a G.P.). First term a=3a=3.
  2. Sum of 10 terms: S10=3(210−1)2−1=3(1024−1)=3(1023)=3069S_{10}=\dfrac{3(2^{10}-1)}{2-1}=3(1024-1)=3(1023)=3069.
  3. Sum of nn terms: Sn=3(2n−1)2−1=3(2n−1)S_n=\dfrac{3(2^n-1)}{2-1}=3(2^n-1).

Part (ii): 1,−3,9,−27,…1,-3,9,-27,\ldots

  1. Here a=1a=1, r=−31=−3r=\dfrac{-3}{1}=-3 (check: 9/(−3)=−39/(-3)=-3 ✓).
  2. Sum of 8 terms: S8=1[(−3)8−1]−3−1=6561−1−4=6560−4=−1640S_8=\dfrac{1\left[(-3)^8-1\right]}{-3-1}=\dfrac{6561-1}{-4}=\dfrac{6560}{-4}=-1640.
  3. Sum of pp terms: Sp=1−(−3)p1−(−3)=1−(−3)p4S_p=\dfrac{1-(-3)^p}{1-(-3)}=\dfrac{1-(-3)^p}{4}.
✓Final answer

(i) S10=3069,Sn=3(2n−1)S_{10}=3069,\quad S_n=3(2^n-1) (ii) S8=−1640,Sp=1−(−3)p4S_8=-1640,\quad S_p=\dfrac{1-(-3)^p}{4}

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.