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Geometric Progression(GP): Formulae based
Example 4
Ex 8.2, 1
Example 5 Important
Ex 8.2, 5 (a)
Ex 8.2, 2
Example 6
Ex 8.2, 4
Ex 8.2, 3 Important You are here
Ex 8.2, 17 Important
Example 7 Important
Ex 8.2, 7 Important
Ex 8.2, 10
Ex 8.2, 9 Important
Ex 8.2, 11 Important
Ex 8.2, 8
Ex 8.2, 19
Ex 8.2, 20
Example 8
Ex 8.2, 13
Ex 8.2, 15
Ex 8.2, 16 Important
Ex 8.2, 21
Ex 8.2, 14 Important
Misc 3
Misc 2
Example 9 Important
Ex 8.2, 12
Example 10 Important
Ex 8.2, 18 Important
Misc 11 (i) Important
Misc 5
Misc 1 Important
Geometric Progression(GP): Formulae based
Last updated at May 29, 2023 by Teachoo
Ex 8.2, 3 The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps. We know that an = arn – 1 where an = nth term of GP n is the number of terms a is the first term r is the common ratio Here, 5th term is p i.e. a5 = p Putting n = 5 in an formula p = ar5–1 p = ar4 Also, 8th term is q i.e. a8 = q Putting n = 8 in an formula q = ar8 – 1 q = ar7 Also, 11th term is s i.e. a11 = s Putting n = 11 in an formula s = ar11 – 1 s = ar10 We need to show that q2 = ps. Taking L.H.S q2 Putting value q = ar7 from (2) = (ar7)2 = a2r14 Now, solving R.H.S ps Putting p = ar4 , s = ar10 from (1)& (3) = (ar4) (ar10) = (a × a)(r10 × r4) = a2r14 = L.H.S Thus, L.H.S = R.H.S Hence proved