The Sum of Nth Term of a G.P

Finding the sum of terms in a geometric progression can be obtained with the use of formulas.

The sum of the first nth term of a G.P is given as 

where	Sn	sum of GP with n terms
	a	the first term
	r	common ratio
	n	number of terms

Sum of a G.P to Infinity:

The sum to infinity of a geometric series,

S_∞ = \( \frac{a}{1 \: – \: r} \) (-1<r<1)

where	S∞	sum of GP with infinitely many terms
	a	the first term
	r	common ratio
	n	number of terms

Example 6.2.1:

If the 2^nd and 5^th terms of a G.P are -6 and 48 respectively, Find the sum of the first four terms. (SSCE)

Solution

T₂ = ar = -6 ……….(1)

T₅ = ar⁴ = 48 ……….(2)

divide equation (2) by equation (1)

⇒ \( \frac {ar^4}{ar} = \frac {48}{-6} \)

r³ = -8

take the cube root of both sides

\(\scriptsize \sqrt[3]{r^3} = \sqrt[3]{-8} \)

r = -2

substitute r = -2 in equation(1)

ar = -6 ……….(1)

\( \scriptsize a \: \times \: (-2) = \: – 6 \)

\( \scriptsize -2a = \: – 6 \)

\( \scriptsize a = \normalsize \frac{-6}{-2} \)

\( \scriptsize a = 3 \)

To find the sum of the first four terms, use the formula:

Note r = -2 which is less than 1

S_n = \( \frac{a(1 \: – \: r^n)}{1 \: – \: r} \) 

a = 3, r = -2

S₄ = \( \frac{3 \left[ (1 \: – \: (-2) ^4 \right]}{1 \: – \: (-2)} \) 

S₄ = \( \frac{3(1 \: – \: 16)}{3} \\ \frac{\not{3}(1 \: – \: 16)}{\not{3}} \\ \scriptsize = -15 \)

Example 6.2.2:

The sum of the first three terms of a G.P. is half its sum to infinity. Find a positive common ratio. (JAMB)

Solution: 

The sum to infinity of a geometric series,

S_∞ = \( \frac{a}{1 \: – \: r} \) (-1<r<1)

S₃ = \( \frac{S_\infty}{2} \)

⇒ \( \frac {ar^3 \: – \: 1}{r \: – \: 1} = \frac{\frac{a}{1 \: – \: r} }{2} \)

⇒ \( \frac {ar^3 \: – \: 1}{r \: – \: 1} = \frac{a}{1 \: – \: r} \: \times \: \frac{1}{2} \)

⇒ \( \frac {ar^3 \: – \: 1}{r \: – \: 1} = \frac {a}{2(1 \: – \: r)} \)

multiply both sides by -1 

⇒ \( \frac{a}{1 \: – \: r} \scriptsize \: \times \: \left (r^3 \: – \:1 \right) = \; – \normalsize \frac {a}{2(1 \: – \: r)} \)

÷ both sides by \( \frac{a}{1 \: – \: r} \)

r³ – 1 = \(\: – \frac{1}{2}\)

r³ = 1 – \( \frac{1}{2}\)

r³ = \( \frac{1}{2}\)

r = \( \sqrt [3]{\frac{1}{2}}\)

Example 6.2.3:

The first term of a G.P. is twice its common ratio. Find the sum of the first two terms of the progression, if its sum to infinity is 8. (JAMB)

Solution: 

a = 2r ………(1)

S_∞ = 8 = \( \frac{a}{1 \: – \: r}\)

8(1 – r) = a ………(2)

substitute for 2r in eqn (2)

8(1 – r) = 2r

÷ both sides by 2

4(1 – r) = r

4 – 4r = r

5r = 4

r = \( \frac{4}{5} \)

substitute for r in eqn(1)

a = 2 × \( \frac{4}{5} \)

a = \( \frac{8}{5} \)

Using S_n = \( \frac{a(1 \: – \: r^n)}{1 \: – \: r} \)  (r < 1)

S₂ = \( \frac{\large \frac{8}{5}\left(\normalsize 1 \: – \: r^2\right)}{\large \frac{1}{1} \: – \: \frac{4}{5} } \\ = \frac{8 \left[ 1 \: – \: \left(\large \frac{4}{5}\right)^2 \right]}{5 \left(\large \frac{5}{5} \: – \: \frac{4}{5} \right)} \)

S₂ = \( \frac{8 \left( \large \frac{1}{1} \: – \: \frac{16}{25}\right)}{5 \left( \large \frac{1}{5}\right) } \)

S₂ = \( \frac {8 \left ( \large\frac{25}{25} \: – \: \frac{16}{25} \right)}{1} \)

S₂ = \( \scriptsize 8 \: \times \: \frac{9}{25} \\ = \frac{72}{25} \)

S₂ = \( \scriptsize 2 \frac{22}{25} \)

Example 6.2.4:

The third term of a G.P is 360 and the 6^th term is 1215. Find the 

i. common ratio
ii. first term
iii. sum of the first four terms (SSCE)

Solution: 

i. Find the common ratio:

The third term:

T₃ = ar^3-1 = ar² = 360

ar² = 360 ……..(1)

The sixth term:

T₆ = ar^6-1 = ar⁵ = 1215

ar⁵ = 1215 ……..(2)

÷ eqn (2) by eqn(1) 

⇒ \( \frac {ar^5}{ar^2} = \frac {1215}{360} \)

⇒ \( \frac {\not{a}r^5}{\not{a}r^2} = \frac {1215}{360} \)

⇒ \( \scriptsize r^{5\:-\:2} = \normalsize \frac {1215}{360} \)

⇒ \( \scriptsize r^{3} = \normalsize \frac {27}{8} \)

⇒ \( \scriptsize r = \normalsize \frac {3}{2} \)

⇒ \( \scriptsize r = 1 \frac {1}{2} \)

ii. Calculate the first term:

Substitute r in equation(1)

a × \( \left (\frac{3}{2} \right)^2 \scriptsize = 360 \)

a × \( \frac{9}{4} \scriptsize = 360 \)

a = \( \frac{360 \: \times \: 4}{9} \)

a = 160

∴ first term = 160

iii. Calculate the sum of the first four terms:

Use the formula: S_n = \( \frac{a(r^n \: – \: 1)}{r \: – \: 1} \)

a = 160, r = \(\frac{3}{2} \)

S₄ = \( \frac{160 \left [\left ( \large \frac{3}{2} \right) ^4 \: – \: 1 \right] }{\large \frac{3}{2} \: – \: 1} \)

S₄ = \( \frac{160 \left ( \large \frac{81}{16} \: – \: \frac{1}{1} \right) }{\large \frac{3}{2} \: – \: \frac{2}{2} } \)

S₄ = \( \frac{160 \left ( \large \frac{81}{16} \: – \: \frac{1}{1} \right) }{\large \frac{1}{2}} \)

S₄ = \( \scriptsize 2 \: \times \: 160 \left ( \normalsize \frac{81 \: – \: 16}{16} \right) \)

= 320 \( (\frac{65}{16})\)

= 20 × 65

S₄ = 1300

The sum of the first four terms is 1300

Example 6.2.5:

If 7 and 189 are the first and fourth terms of a geometric progression respectively, find the sum of the first three terms of the progression. (JAMB).

Solution:

a = 7 ….(i)

T₄ = ar^4-1 = ar³ = 189 …..(ii)

substitute a in equation (i) into equation (ii)

ar³ = 189 …..(ii)

(7)r³ = 189

r³ = \( \frac{189}{7} \)

r³ = 27

take the cube root of both sides

\( \scriptsize \sqrt[3]{r^3} = \sqrt[3]{27} \)

r = \(\scriptsize \sqrt[3] {27} = 3 \)

To find ind the sum of the first three terms we use the formula:

S_n = \( \frac{a(r^n \: – \: 1)}{r \: – \: 1} \)

a = 7, r = 3

S₃ = \( \frac{7(3^3 \:-\: 1)}{3 \: – \: 1} \)

= \( \frac{7(27 \: – \: 1)}{2} \)

= \( \frac{7(26)}{2} \)

= 7 × 13

S₃ = 91

∴ The sum of the first three terms of the progression = 91

Exercise: