Mathematics Class 11
🏠 Home Log in
TOPIC 25 OF 45

8.3.1 Sum of n terms of a G.P.

🎓 Class 11 Mathematics CBSE Theory Ch 8 — Sequences and Series ⏱ ~15 min
🌐 Language: [gtranslate]

This MCQ module is based on: 8.3.1 Sum of n terms of a G.P.

This mathematics assessment will be based on: 8.3.1 Sum of n terms of a G.P.
Targeting Class 11 level in Sequences Series, with Advanced difficulty.

Upload images, PDFs, or Word documents to include their content in assessment generation.

8.3.1 Sum of n terms of a G.P.

Sum formula
The sum \(S_n=a+ar+ar^2+\cdots+ar^{n-1}\) of the first \(n\) terms of a G.P. with first term \(a\) and common ratio \(r\) is: \[\boxed{\;S_n=\dfrac{a(r^n-1)}{r-1}\quad (r\ne 1)\;}\] Equivalently \(\dfrac{a(1-r^n)}{1-r}\), preferred when \(r<1\) so the denominator stays positive. If \(r=1\), every term is \(a\), so \(S_n=na\).
Proof (the "shift trick")
Write \(S_n=a+ar+ar^2+\cdots+ar^{n-1}\) (eqn. 1).
Multiply by \(r\): \(rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n\) (eqn. 2).
Subtract (1) from (2): \(rS_n-S_n=ar^n-a\), i.e. \((r-1)S_n=a(r^n-1)\). Divide by \(r-1\) (since \(r\ne 1\)) to get the formula. \(\square\)

8.3.2 Sum of an Infinite G.P.

If \(|r|<1\), the term \(r^n\) shrinks to zero as \(n\to\infty\). Substituting in the finite formula:

Infinite-sum formula
For \(|r|<1\): \[\boxed{\;S_\infty=\dfrac{a}{1-r}\;}\] For \(|r|\ge 1\), the terms don't approach zero, so the infinite sum diverges (does not exist as a finite number).

This is the foundation of the geometric-series convergence theorem and underpins all of compound-interest, present-value calculation, and the geometric-distribution generating function in probability.

8.4 Geometric Mean (G.M.)

Geometric Mean
The geometric mean of two positive numbers \(a\) and \(b\) is \[\boxed{\;G=\sqrt{ab}\;}\] It is the middle term of a 3-term G.P. with first term \(a\) and last term \(b\) (so the sequence is \(a, \sqrt{ab}, b\)).
For \(n\) numbers \(a_1, a_2, \ldots, a_n\) the geometric mean is \(\sqrt[n]{a_1a_2\cdots a_n}\).

Inserting G.M.s between two numbers

To insert \(k\) geometric means \(G_1, G_2, \ldots, G_k\) between \(a\) and \(b\) (so \(a, G_1, \ldots, G_k, b\) is a G.P. of length \(k+2\)):

\[r=\left(\dfrac{b}{a}\right)^{1/(k+1)},\qquad G_i=ar^i.\]

8.4.1 Relation between A.M. and G.M.

AM–GM inequality
For positive reals \(a, b\): \[\dfrac{a+b}{2}\ge\sqrt{ab},\quad\text{i.e. } A\ge G,\] with equality iff \(a=b\).

Proof: \((\sqrt a-\sqrt b)^2\ge 0\Rightarrow a-2\sqrt{ab}+b\ge 0\Rightarrow a+b\ge 2\sqrt{ab}\). \(\square\)

This is the simplest case of the celebrated AM–GM inequality, generalising to \(n\) numbers and lying at the heart of optimisation, statistics, and information theory.

8.5 Special Series — Sum of First n Natural Numbers, Squares, Cubes

Three formulas to memorise
\[\sum_{k=1}^{n}k=\dfrac{n(n+1)}{2}\quad(\text{the triangular numbers})\] \[\sum_{k=1}^{n}k^2=\dfrac{n(n+1)(2n+1)}{6}\] \[\sum_{k=1}^{n}k^3=\left[\dfrac{n(n+1)}{2}\right]^2=\!\left(\sum_{k=1}^{n}k\right)^{\!2}\] The cubes' sum is the SQUARE of the linear sum — a beautiful identity due to Aryabhata.

Worked Examples

Example 9. Find the sum of \(2+6+18+54+\cdots\) up to 8 terms.
G.P. with \(a=2,\ r=3,\ n=8\). \(S_8=\dfrac{2(3^8-1)}{3-1}=\dfrac{2\cdot 6560}{2}=6560\).
Example 10. The sum of first three terms of a G.P. is 13/12 and their product is −1. Find the common ratio and the terms.
Let the three terms be \(a/r,\ a,\ ar\). Their product = \(a^3=-1\Rightarrow a=-1\). Their sum: \(-1/r-1-r=13/12\), giving \(12r^2+25r+12=0\). Solve: \(r=-3/4\) or \(-4/3\). Two G.P.s: \((4/3,-1,3/4)\) for \(r=-3/4\), or \((3/4,-1,4/3)\) for \(r=-4/3\).
Example 11. Find the sum of \(7+77+777+7777+\cdots\) up to n terms.
\(S_n=7\cdot[1+11+111+\cdots]=\dfrac{7}{9}\cdot[9+99+999+\cdots]=\dfrac{7}{9}[(10-1)+(100-1)+(1000-1)+\cdots]\) \(=\dfrac{7}{9}\!\left[\dfrac{10(10^n-1)}{10-1}-n\right]=\dfrac{7}{9}\!\left[\dfrac{10(10^n-1)}{9}-n\right]=\dfrac{7}{81}[10(10^n-1)-9n]\).
Example 12. Insert 3 geometric means between 1 and 256.
\(a=1,\ b=256,\ k=3\). \(r=(256/1)^{1/4}=4\). G.M.s: \(1\cdot 4=4\), \(1\cdot 16=16\), \(1\cdot 64=64\). G.P.: 1, 4, 16, 64, 256 ✓.
Example 13. If A and G are A.M. and G.M. of two positive numbers, prove the numbers are \(A\pm\sqrt{A^2-G^2}\).
Let the numbers be \(a,b\). Then \(a+b=2A\) and \(ab=G^2\). So \(a,b\) are roots of \(t^2-2At+G^2=0\). By the quadratic formula: \(t=A\pm\sqrt{A^2-G^2}\). \(\square\)
Example 14. Find the sum \(1+2\cdot 2+3\cdot 4+4\cdot 8+\cdots+n\cdot 2^{n-1}\) (an "arithmetic-geometric" series).
\(S=\sum_{k=1}^{n}k\cdot 2^{k-1}\). Use the shift trick: \(2S-S=\sum k\cdot 2^k-\sum k\cdot 2^{k-1}\). Re-arranging gives \(S=(n-1)\cdot 2^n+1\).
Check \(n=1\): \(S_1=1\), formula gives \(0\cdot 2+1=1\) ✓. \(n=3\): \(1+4+12=17\), formula \(2\cdot 8+1=17\) ✓.
Example 15. Find \(\sum_{k=1}^{20} k^3\).
\([20\cdot 21/2]^2=[210]^2=44100\).
Activity: Convert a Repeating Decimal to a Fraction
L4 Analyse
Materials: Calculator, paper.
Predict: What fraction equals 0.333333… ? Use the infinite G.P. formula.
  1. Write 0.333… = 0.3 + 0.03 + 0.003 + … = 3/10 + 3/100 + 3/1000 + … .
  2. This is an infinite G.P. with \(a=3/10\), \(r=1/10\). Since \(|r|<1\), use \(S_\infty=a/(1-r)\).
  3. \(S_\infty=(3/10)/(9/10)=3/9=1/3\). ✓
  4. Now try 0.142857142857… (the decimal of 1/7): \(a=142857/10^6,\ r=1/10^6\). \(S_\infty=142857/999999=1/7\). ✓
  5. Conclusion: every repeating decimal is a fraction (rational), and the geometric-series formula gives the conversion in one line.
All repeating decimals are rationals; conversely all rationals have either terminating or eventually-repeating decimals (by the pigeonhole principle on remainders). The geometric-series sum is the cleanest tool for this conversion. Irrationals like \(\pi\) and \(\sqrt 2\) have non-repeating, non-terminating decimal expansions.

Competency-Based Questions

Scenario: A bank pays 8% annual compound interest. ₹10,000 is deposited.
Q1. The amount at the end of 5 years is:
L3 Apply
Answer: \(10000\cdot 1.08^5\approx 10000\cdot 1.4693=₹14693\).
Q2. The sum of the infinite G.P. \(2 + 1 + 1/2 + 1/4 + \cdots\) is:
L3 Apply
  • (a) 3
  • (b) 4
  • (c) 5
  • (d) ∞
Answer: (b) 4. a = 2, r = 1/2, |r| < 1: \(S_\infty=2/(1-1/2)=4\).
Q3. (T/F) "For positive a, b: AM = GM iff a = b." Prove.
L5 Evaluate
True. AM − GM = (a+b)/2 − √(ab) = (√a − √b)²/2 ≥ 0, with equality iff √a = √b iff a = b.
Q4. Compute \(\sum_{k=1}^{15} k^2\) and \(\sum_{k=1}^{15} k^3\).
L3 Apply
Solution: \(\sum k^2 = 15\cdot 16\cdot 31/6=1240\). \(\sum k^3=(15\cdot 16/2)^2=120^2=14400\).
Q5. Design: a ball is dropped from height 8 m and rebounds to 3/4 of its previous height each bounce. Find the total vertical distance travelled before it stops bouncing.
L6 Create
Solution: First fall: 8. Then up 6 + down 6 = 12. Then up 4.5 + down 4.5 = 9. … Total = 8 + 2(6 + 4.5 + 3.375 + …) = 8 + 2·(6/(1−3/4)) = 8 + 2·24 = 56 m.

Assertion–Reason Questions

Assertion (A): The sum of the infinite G.P. \(1 + 2 + 4 + 8 + \cdots\) does not exist as a finite number.
Reason (R): The infinite-sum formula \(a/(1-r)\) requires \(|r| < 1\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). r = 2 fails the |r| < 1 requirement; partial sums grow without bound.
Assertion (A): \(1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2\).
Reason (R): Both sides equal \([n(n+1)/2]^2\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). R is the common closed-form value of both sides.
Assertion (A): If a and b are positive real numbers with \(a+b=10\) and \(ab=21\), then a, b are 7 and 3.
Reason (R): a, b are roots of \(t^2-10t+21=0\) (Vieta's relations), giving \(t=3\) or \(7\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). R is the standard quadratic-from-AM-and-product trick (Example 13's idea).

Frequently Asked Questions

What is the formula for the sum of n terms of a G.P.?
Sₙ = a(rⁿ − 1)/(r − 1) for r ≠ 1. If r = 1, Sₙ = na.
What is the sum of an infinite G.P. with |r| < 1?
S∞ = a/(1 − r). Example: 1 + 1/2 + 1/4 + … = 2.
What is the geometric mean of two numbers?
G.M. of positive numbers a and b is √(ab). It is the middle term of a 3-term G.P. with first term a and last term b.
What is the AM-GM inequality?
For positive reals a, b: (a + b)/2 ≥ √(ab), with equality iff a = b. Proof: (√a − √b)² ≥ 0.
What is the sum of the first n natural numbers?
1 + 2 + 3 + … + n = n(n+1)/2.
What is the sum of the squares of the first n natural numbers?
1² + 2² + … + n² = n(n+1)(2n+1)/6. The cubes' sum is [n(n+1)/2]² — the square of the linear sum.
AI Tutor
Mathematics Class 11
Ready
Hi! 👋 I'm Gaura, your AI Tutor for 8.3.1 Sum of n terms of a G.P.. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.