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CAT 2025 Lesson : Progressions - GP Concepts

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3. Geometric Progression


If the ratio of any two consecutive terms of a sequence is the same constant, then the sequence is said to be in Geometric Progression (GP).

If x1,x2,x3,....,xnx_1,x_2,x_3,...., x_nare in Geometric Progression, then Common Ratio

= x2x1\dfrac{x_2}{x_1}=x3x2\dfrac{x_3}{x_2} = x4x3\dfrac{x_4}{x_3} = ..... = xnxn−1\dfrac{x_n}{x_{n-1}}

Where a\bm{a} is the first term and r\bm{r} is the common ratio, the GP is of the form a, ar, ar2ar^{2},ar3ar^{3}, ... , arn−1ar^{n-1}.

3.1 Terms & Formulae for GP


Where aa and rr are the first term and common ratio respectively in a GP with nn terms,

1) nth\bm{n^{th}} term = arn−1ar^{n-1}

2) Geometric Mean of GP = GM of First and Last terms = x1×xn\sqrt{x_1 \times x_n}

= a×ar(n−1)\sqrt{a \times ar^{(n-1)}} = ar(n−1)/2 ar^{(n - 1)/2}

3) Sum of n terms of a GP = Sn\bm{S_n} = a(rn−1)r−1\dfrac{a(r^{n}-1)}{r-1}

4) If a GP has infinite terms and 0<r<1 0 \lt r \lt 1, then Sum of infinite terms of the GP = a(r−1)\dfrac{a}{(r-1)}

3.2 Properties of GP


1) If each term is multiplied or divided by a constant, then the resulting sequence is also in GP.

2) In two GPs with the same number of terms, if the corresponding terms by position in the two GPs are multiplied or divided, then the resulting sequence will also be in GP.

3) Geometric mean is also the middle term where the number of terms is odd or the geometric mean of the two middle terms where the number of terms is even.

4) If a a, bb and cc are in GP, then bb = ac\sqrt{ac}

5) The reciprocals of the terms of a GP also form a GP, wherein, the common ratio of the new GP is the reciprocal of the common ratio of the earlier GP.

Example 12

What is the sum of the first 66 terms of the GP 2,6,182, 6, 18, ... ?

Solution

a=2a = 2, r=62=3r =\dfrac{6}{2} = 3, n=6n = 6

Sum of 66 terms = a(rn−1)(r−1)=2(36−1)(3−1)=728\dfrac{a(r^{n}-1)}{(r-1)} = \dfrac{2(3^{6}-1)}{(3-1)} = 728

Answer: 728728

Example 13

In an increasing GP, the difference between the first and third terms is 9696 and the difference between the second and fourth terms is 480480. What is the first term of this GP?

Solution

Let the 44 terms be aa,arar, ar2ar^{2} and ar3ar^{3}

⇒ ar2−a=96ar^{2}- a = 96-----(1)

⇒ ar3−ar=480ar^{3}- ar =480

⇒ r(ar2−a)=480r(ar^{2}- a)= 480-----(2)

When (2)(2) is divided by (1)(1) ⇒ r(ar2−a)ar2−a=48096\dfrac{r(ar^{2}-a)}{ar^{2}-a} = \dfrac{480}{96}

⇒ r=5 r = 5

Substituting in (1) ⇒ 25a25a – aa = 9696

⇒ aa = 4\bm{4}

Answer: 44

Example 14

What is the sum of all terms in the GP 4,43,49,427,....4,\dfrac{4}{3},\dfrac{4}{9},\dfrac{4}{27},....?

Solution

a=4a = 4, r=13r =\dfrac{1}{3}

As 0<r<1 0 \lt r \lt 1, we can apply the sum to infinity formula.

Sum of infinite terms in GP = a1−r\dfrac{a}{1- r} = 4(1−13)\dfrac{4}{\left( 1-\dfrac{1}{3} \right)} = 4×32=64 \times \dfrac{3}{2} = 6

Answer: 66

Example 15

If the sum of first 22 terms of a decreasing GP is 99, and the sum of infinite terms in this GP is 1212, then what is the common ratio for this GP?

Solution

a+ar=9a + ar = 9 ⇒ a=91+ra =\dfrac{9}{1+r} -----(1)

a1−r=12\dfrac{a}{1 - r}= 12 ⇒ aa = 12(1 – rr) -----(2)

Equating (1)(1) and (2)(2) ⇒ 91+r=12(1−r)\dfrac{9}{1+r} = 12(1 - r)

⇒ 1−r2=341 - r^{2} = \dfrac{3}{4} ⇒ r2=14r^{2} = \dfrac{1}{4} ⇒ r=12r = \dfrac{1}{2} [As this is an infinite decreasing GP, rr = is rejected]

Answer: 12\dfrac{1}{2}

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