WebAnswer (1 of 3): GP T8 =ar^7 =384 T3=ar^2=12 T8/T3= r^5 =384 /12 =32 r^5=2^5 r=2 T3 =ar^2=a ×4=12 a=12/4=3 a=3 r=2 S10 =a(r^n --1)/(r-1) =3(2^10 -1) /(2-1) = 3×(2^10 -1) 3×(1024-1) =3069 Sum of 10 terms of GP is 3069 WebJun 30, 2024 · in a G.P,the sum of the first and the last term is 66,the product of the second and last but one term is 128 and the sum of the terms is 126. [a] if an increasing G.P is …

in a G P,the sum of the first and the last term is 66,the product of ...

WebThe geometric sequence formula to determine the sum of the first n terms of a Geometric progression is given by: S_n = a [ (r^n-1)/ (r-1)] if r > 1 and r ≠ 1 S_n = a [ (1 – r^n)/ (1 – r)] if r < 1 and r ≠ 1 The nth item at the end of GP, the last item is l, … WebJun 26, 2024 · Find the sum of all the terms, if the first $3$ terms among $4$ positive $2$ digit integers are in AP and the last $3$ terms are in GP. Moreover the difference between the first and last term is 40. Moreover the difference between the first and last term is 40. the payment has not been received

In an increasing G.P , the sum of the first and the last …

WebIn an increasing gp the sum of the first and the last term is 66. The product of the second and the last but one is 128 and the sum of the sum of the terms is 126 ..then the no. Of … WebSep 2, 2024 · Identify the first and last terms in the sequence. You need to know both of these numbers in order to calculate the sum of the arithmetic sequence. Often the first numbers will be 1, but not always. Let the variable equal the first term in the sequence, and equal the last term in the sequence. WebJun 20, 2024 · n=6 terms (ii)sum of 'n' terms in GP is given by. S=a(r^n-1)/(r-1) S=3(2^6-1)/(2-1) S=3(64-1)/1. S=3(63) S=189. 3,6,12,24,48,96 are the numbers that are in GP. Advertisement Advertisement Ritiksuglan Ritiksuglan Answer: (i)given first term(a)=3. last term(T)=96. common ratio(r)=2. last term in GP is ar^(n-1),n is total number of … shy ministers hide one old database