Sum of n Terms of Geometric Progression

The sum of n terms of a geometric progression, denoted by [Sₙ], means the total obtained by adding the first [n] terms of the G.P. Instead of adding each term individually, we use a formula to calculate this sum efficiently.

To find [Sₙ], we must know the first term [a], the common ratio [r], and the number of terms [n]. Without any one of these, the sum cannot be uniquely determined.

When [r ≠ 1], the sum of first [n] terms is given by [ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]. This formula is valid for all real values of [r] except 1.

When [r = 1], all terms of the G.P. become equal to [a]. The general formula leads to division by zero, which is undefined. Hence, we use the simple formula [Sₙ = na].

The formula is derived by writing the sum [Sₙ], multiplying it by [r], and subtracting the two expressions. This subtraction causes most terms to cancel, leaving only the first and last terms.

Yes. The formula works perfectly for fractional values of [r], such as [\dfrac{1}{2}] or [\dfrac{1}{3}], provided [r ≠ 1].

Yes. If the first term [a] is negative, or if the terms alternate due to a negative [r], the sum can be negative.

No. When [0 < r < 1], each new term added is smaller than the previous one, so the increase in sum slows down.

Yes. In some G.P.s with negative common ratio, positive and negative terms may cancel each other, making the sum zero.

When [r] is negative, the terms alternate in sign. As a result, the sum may increase or decrease irregularly depending on the number of terms.

Yes. If [Sₙ], [a], and [n] are known, the formula can be rearranged to find the common ratio [r].

Yes. By substituting known values and solving the resulting equation, we can find [n].

Yes. The first term must be correctly identified as [a]; otherwise, the sum will be incorrect.

It is used in compound interest, population growth, radioactive decay, and financial calculations.

The sum of a G.P. depends on multiplication through the common ratio, whereas the sum of an A.P. depends on addition through the common difference.

Incorrect. When [r = 1], division by zero occurs. We must use [Sₙ = na].

False. When [0 < r < 1], the sum increases slowly and may remain small.

Incorrect. Fractional values of [r] are perfectly valid.

False. Negative [r] simply causes alternating signs.

Incorrect. A.P. and G.P. have completely different sum formulas.

Not always. It depends on the value of [r].

Incorrect. The first term must be identified correctly.

Incorrect. Certain G.P.s can have zero sum.

False. It works for both increasing and decreasing G.P.s.

Incorrect. They have wide practical applications in science and finance.

Step-by-Step Solution:

[a = 1]

[r = 2]

[n = 7]

[S₇ = \dfrac{1(2⁷ − 1)}{1}]

[2⁷ = 128]

[S₇ = 127]

Answer: 127

Step-by-Step Solution:

[a = 3]

[r = 3]

[n = 5]

[S₅ = \dfrac{3(3⁵ − 1)}{2}]

[3⁵ = 243]

[S₅ = 363]

Answer: 363

Step-by-Step Solution:

First term, [a = 3]

Common ratio, [r = \dfrac{6}{3} = 2]

Number of terms, [n = 6]

Formula for sum of n terms (since [r ≠ 1]):
[ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]

Substitute the values:
[ S₆ = \dfrac{3(2⁶ − 1)}{2 − 1} ]

Calculate the power:
[ 2⁶ = 64 ]

Simplify:
[ S₆ = \dfrac{3(64 − 1)}{1} = 3 × 63 ]

Final answer:
[ S₆ = 189 ]

Step-by-Step Solution:

First term, [a = 2]

Common ratio, [r = \dfrac{6}{2} = 3]

Number of terms, [n = 5]

Use the formula:
[ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]

Substitute values:
[ S₅ = \dfrac{2(3⁵ − 1)}{3 − 1} ]

Calculate power:
[ 3⁵ = 243 ]

Simplify:
[ S₅ = \dfrac{2(243 − 1)}{2} = 242 ]

Final answer:
[ S₅ = 242 ]

Step-by-Step Solution:

First term, [a = 1]

Common ratio, [r = 2]

Number of terms, [n = 8]

Formula:
[ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]

Substitute:
[ S₈ = \dfrac{1(2⁸ − 1)}{2 − 1} ]

Calculate power:
[ 2⁸ = 256 ]

Simplify:
[ S₈ = 256 − 1 ]

Final answer:
[ S₈ = 255 ]

Step-by-Step Solution:

First term, [a = 5]

Common ratio, [r = \dfrac{10}{5} = 2]

Number of terms, [n = 7]

Apply formula:
[ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]

Substitute values:
[ S₇ = \dfrac{5(2⁷ − 1)}{1} ]

Calculate power:
[ 2⁷ = 128 ]

Simplify:
[ S₇ = 5(128 − 1) = 5 × 127 ]

Final answer:
[ S₇ = 635 ]

Step-by-Step Solution:

First term, [a = 81]

Common ratio, [r = \dfrac{27}{81} = \dfrac{1}{3}]

Number of terms, [n = 4]

Formula:
[ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]

Substitute:
[ S₄ = \dfrac{81\left((\dfrac{1}{3})⁴ − 1\right)}{\dfrac{1}{3} − 1} ]

Calculate power:
[ (\dfrac{1}{3})⁴ = \dfrac{1}{81} ]

Simplify expression:
[ S₄ = \dfrac{81(\dfrac{1}{81} − 1)}{-\dfrac{2}{3}} ]

Final answer:
[ S₄ = 120 ]

Step-by-Step Solution:

First term, [a = 4]

Common ratio, [r = \dfrac{-8}{4} = -2]

Number of terms, [n = 6]

Formula:
[ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]

Substitute values:
[ S₆ = \dfrac{4((-2)⁶ − 1)}{-2 − 1} ]

Calculate power:
[ (-2)⁶ = 64 ]

Simplify:
[ S₆ = \dfrac{4(64 − 1)}{-3} ]

Final answer:
[ S₆ = -84 ]

Step-by-Step Solution:

First term, [a = 1]

Common ratio, [r = 3]

Number of terms, [n = 10]

Use formula:
[ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]

Substitute:
[ S₁₀ = \dfrac{3¹⁰ − 1}{2} ]

Calculate power:
[ 3¹⁰ = 59049 ]

Simplify:
[ S₁₀ = \dfrac{59048}{2} ]

Final answer:
[ S₁₀ = 29524 ]

Step-by-Step Solution:

First term, [a = 16]

Common ratio, [r = \dfrac{8}{16} = \dfrac{1}{2}]

Number of terms, [n = 5]

Formula:
[ Sₙ = \dfrac{a(rⁿ − 1)}{r − 1} ]

Substitute values:
[ S₅ = \dfrac{16((\dfrac{1}{2})⁵ − 1)}{\dfrac{1}{2} − 1} ]

Calculate power:
[ (\dfrac{1}{2})⁵ = \dfrac{1}{32} ]

Simplify expression:
[ S₅ = \dfrac{16(\dfrac{1}{32} − 1)}{-\dfrac{1}{2}} ]

Final answer:
[ S₅ = 31 ]