Ex 8.2

Chapter 8 Class 11 Sequences and Series
Serial order wise

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### Transcript

Ex 9 .3, 14 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P. Given Now, Sn = (๐(1 โ ๐^๐))/(1 โ ๐ ) where Sn = sum of n terms of GP n is the number of terms a is the first term r is the common ratio We know that S3 = 16 Putting n = 3 in Sn (๐(1 โ ๐3 ))/(1 โ ๐ ) = 16 Also, S6 = 144 Putting n = 6 in Sn (a(1 โ r6))/(1โ r) = 144 We need to find first term (a) & common ratio (r) Dividing (2) from (1) ((๐(1 โ ๐6))/(1โ ๐))/((๐(1 โ ๐3))/(1 โ ๐)) = 144/16 (๐(1 โ ๐6))/(1โ ๐) ร (1โ ๐)/(๐(1 โ ๐3)) = 144/16 (๐(1 โ ๐6)(1โ ๐))/(๐(1 โ ๐3)(1โ ๐)) = 9/1 (1 โ๐6)/(1 โ๐3) = 9/1 (12 โ(๐3)2)/(1 โ ๐3) = 9/1 Using a2 โ b2 = (a โ b)(a + b) ((1 โ ๐3)(1 + ๐3))/((1 โ ๐3)) = 9/1 ((1+ ๐3))/1 = 9/1 ((1+ ๐3))/1 = 9/1 1 + r3 = 9 r3 = 9 โ 1 r3 = 8 r =โ8 r = โ(2 ร2 ร2) r = 2 Hence common ratio r = 2 Putting r = 2 in (1) (๐(1 โ๐3))/(1 โ๐) = 16 (๐(1 โ23))/(1 โ2) = 16 (๐(1 โ8))/(โ1) = 16 (๐(1 โ8))/(โ1) = 16 (๐(โ7))/(โ1) = 16 a(7) = 16 a = 16/7 Hence first term (a) =16/7 Now, a = 16/7 , r = 16/7 Now we need to find sum of n terms As r > 1 Sum of n terms of G.P = (๐(๐^๐ โ 1))/(๐ โ 1) Sum of n terms of G.P = (๐(๐^๐ โ 1))/(๐ โ 1) Putting a = 16/7 & r = 2 Sn = (16/7(2๐ โ 1 ))/(2 โ 1 ) = (16/7(2๐ โ 1 ))/1 = 16/7 (2n โ 1) Hence sum of n terms is 16/7 (2n โ 1)