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CAT 2025 Lesson : Progressions - GP Special Types

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3.3 GP where n is small

In GP with a small number of terms, you can use the following format for ease in calculations.

1) Where the number of terms is odd , take the
xxx as the middle term and yyy as the common ratio.

- For a
333-term GP, use the terms xy,x,xy\dfrac{x}{y},x,xyyx​,x,xy

- For a
555-term GP, use the terms xy2,xy,x,xy,xy2\dfrac{x}{y^{2}},\dfrac{x}{y},x,xy,xy^{2}y2x​,yx​,x,xy,xy2

2) Where the number of terms is even, take
xy\dfrac{x}{y}yx​ and xy xy xy as middle terms and y2y^{2}y2 as the common ratio.

- For a
444-term GP, use the terms xy3,xy,xy,xy3\dfrac{x}{y^{3}},\dfrac{x}{y},xy,xy^{3}y3x​,yx​,xy,xy3

- For a
666-term GP, use the terms xy5,xy3,xy,xy,xy3,xy5\dfrac{x}{y^{5}},\dfrac{x}{y^{3}},\dfrac{x}{y},xy,xy^{3},xy^{5}y5x​,y3x​,yx​,xy,xy3,xy5

Example 16

If the product of 444 terms of an increasing GP is 129612961296, and the product of the 2nd2^{nd}2ndand 4th4^{th}4thterms is 144144144. Then, what is the 1st1^{st}1stterm?

Solution

Let the 444 terms bexy3,xy,xy,xy3\dfrac{x}{y^{3}},\dfrac{x}{y},xy,xy^{3}y3x​,yx​,xy,xy3

Product of
444 terms = x4=1296=64x^{4} = 1296 = 6^{4}x4=1296=64⇒ x=6x = 6x=6

Product of
2nd2nd 2nd and 4th4th4th terms = xy×xy3=x2y2=144\dfrac{x}{y}\times xy^{3} = x^{2}y^{2} = 144yx​×xy3=x2y2=144

⇒
62×y2=1446^{2} \times y^{2} = 14462×y2=144 ⇒ y=2y = 2y=2

∴\therefore∴ 1st 1^{st}1st term = xy3=68=34\dfrac{x}{y^{3}} = \dfrac{6}{8} = \dfrac{3}{4}y3x​=86​=43​

Answer:
34\dfrac{3}{4}43​

3.4 Inserting Geometric Means


When Geometric Means are inserted between two numbers, say
x x x and y, y, y, then all these numbers together would form a Geometric Progression where x x x and y y y will be the first and last terms respectively.

In any GP with
n n n terms, there are (n–2)(n – 2)(n–2) geometric means between the first and the last terms. No direct formula is required for this. Please logically apply the GP formulae where required.

Example 17

If mmm and nnn are the 222 Geometric Means inserted between 666 and 384384384, then m+n m + n m+n = ?

Solution

As the numbers along with the GMs will be in GP, let the terms be a,ar, a, ar,a,ar, ar2ar^{2}ar2 and ar3ar^{3}ar3 respectively.

1st1^{st}1st term = a=6\bm{a = 6}a=6

4th4^{th}4thterm = ar3=384ar^{3}= 384ar3=384 ⇒ r3=3846r^{3}=\dfrac{384}{6}r3=6384​

⇒
r3=64r^{3}= 64r3=64 ⇒ r=4\bm{r = 4}r=4

m+n=ar+ar2=ar(1+r)=6×4×5m + n = ar + ar^{2} = ar( 1 + r ) = 6 \times 4 \times 5m+n=ar+ar2=ar(1+r)=6×4×5 =120\bm{120}120

Answer:
120120120

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