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Difficulty: MediumArithmetic and Geometric Progressions (AP and GP)

The sum to infinity of a geometric progression (G.P.) with positive terms is 1818, and the sum of its first two terms is 1616. What is the first term of the progression?

Answer: 12

Answer

The first term of the progression is 12.
Using the sum to infinity formula S∞=a1−r=18S_{\infty} = \frac{a}{1 - r} = 18, we express the first term as a=18(1−r)a = 18(1 - r). Combining this with the sum of the first two terms S2=a(1+r)=16S_2 = a(1 + r) = 16 yields 18(1−r)(1+r)=16  ⟹  18(1−r2)=1618(1 - r)(1 + r) = 16 \implies 18(1 - r^2) = 16. Solving for rr gives r2=19r^2 = \frac{1}{9}, so r=13r = \frac{1}{3} for a sequence with positive terms. Substituting r=13r = \frac{1}{3} back into a=18(1−r)a = 18(1 - r) gives a=12a = 12.

Step-by-Step Solution

1
Express the sum to infinity in terms of the first term aa and common ratio rr.
a=18(1−r)a = 18(1 - r)
The sum to infinity formula for a convergent G.P. is S∞=a1−rS_{\infty} = \frac{a}{1 - r}.
2
Write the expression for the sum of the first two terms.
a(1+r)=16a(1 + r) = 16
The sum of the first two terms is T1+T2=a+ar=a(1+r)T_1 + T_2 = a + ar = a(1 + r).
3
Substitute a=18(1−r)a = 18(1 - r) into the sum of the first two terms equation.
18(1−r2)=1618(1 - r^2) = 16
Applying the difference of two squares identity (1−r)(1+r)=1−r2(1 - r)(1 + r) = 1 - r^2.
4
Solve for the common ratio rr.
r=13r = \frac{1}{3}
Rearranging gives 1−r2=89  ⟹  r2=191 - r^2 = \frac{8}{9} \implies r^2 = \frac{1}{9}. Since all terms are positive, rr must be positive.
5
Calculate the first term aa.
a=12a = 12
Substitute r=13r = \frac{1}{3} into a=18(1−r)a = 18(1 - r) to get a=18×23=12a = 18 \times \frac{2}{3} = 12.

Key Concept

Geometric Progression sum to infinity and partial sums
Estimated Time:1m 30s
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Topics

Arithmetic and Geometric Progressions (AP and GP)

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