Ex 8.2, 19 - Chapter 8 Class 11 Sequences and Series
Last updated at Dec. 16, 2024 by Teachoo
Geometric Progression(GP): Formulae based
Ex 8.2, 6
You are here
You are here
Chapter 8 Class 11 Sequences and Series
Concept wise
- Finding Sequences
- Arithmetic Progression (AP): Formulae based
- Arithmetic Progression (AP): Statement
- Arithmetic Progression (AP): Calculation based/Proofs
- Inserting AP between two numbers
- Arithmetic Mean (AM)
-
Geometric Progression(GP): Formulae based
- Geometric Progression(GP): Statement
- Geometric Progression(GP): Calculation based/Proofs
- Inserting GP between two numbers
- Geometric Mean (GM)
- AM and GM (Arithmetic Mean And Geometric mean)
- AP and GP mix questions
- Finding sum from series
- Finding sum from nth number
Transcript
Ex 8.2, 19 Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2 1st sequence is 2, 4, 8, 16, 32 2nd sequence is 128, 32, 8, 2, 1/2 Let P = Sum of product of corresponding terms of 1st & 2nd sequences. P = 2 128 + 4 32 + 8 8 + 16 2 + 32 1/2 P = 256 + 128 + 64 + 32 + 16 P = 256 + 128 + 64 + 32 + 16 Now, in 256 , 128 , 64 , 32 ,16 128/256 = 1/2 & 64/128 = 1/2 Thus, ( )/( ) = ( )/( ) i.e. common ratio is same Thus, it is a G.P Here, First term, a = 256 & common ratio r = 128/256 & n = 5 We need to find sum of these terms We need to find sum of these terms Sn = (a(1 ^ ))/(1 r) where Sn = sum of n terms of GP n is the number of terms a is the first term r is the common ratio Putting n = 5 , a = 256 , r = 1/2 S5 = (256[1 (1/2)^5])/(1 1/2) = (256[1 (1/32)])/(1/2) = 256[1 (1/32)] 2 = 256((32 1)/32) 2 = 256((32 1)/32) 2 = (256 31 2)/32 = 16 31 = 496 Hence the sum of products of corresponding terms in sequences is 496
