WebThe \( 6^{\text {th }} \) term of a \( \mathrm{GP}\) is \(32\) and its \( 8^{\text {th }} \) term is \(128\) , then the common ratio of the \( \mathrm{GP}\) ... WebThe \( 6^{\text {th }} \) term of a \( \mathrm{GP}\) is \(32\) and its \( 8^{\text {th }} \) term is \(128\) , then the common ratio of the \( \mathrm{GP}\) ...

Nth term of GP Geometric Progression Solved Examples - Cuemath

WebJul 31, 2024 · The 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94. Find the common difference of the AP. asked Apr 15, 2024 in Arithmetic Progression by Vevek01 ( 47.4k points) WebAnswer (1 of 4): First assume common ratio is r and first term as a. T3=a*r^2=108 T6= a*r^5 =32 T6/T3=32/108=8/27=r^3 Therefore r=2/3 a= 108*9/4=243 Sum of first 7 terms S7=243 [1-(2/3)^7]/(1/3) =729(1–128/3*729) =729-128/3 =729–(42+2/3) =686and 1/3 jmrd wealth management

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WebJul 28, 2024 · Second last term in the GP = a n-1 = ar n-2. Starting from the end, the series forms another GP in the form, ar n-1, ar n-2, ar n-3 ….ar 3, ar 2, ar, a → (2) Common ratio of this GP is \(\frac{1}{r}\) So, common ratio = 2. a = \(\frac{1}{1024}\) So, 6 th term of the GP (2), a 6 = ar 5 = \(\frac{1}{1024}\) x 2 5 = \(\frac{1}{32}\) Hence, the ... WebThe Question and answers have been prepared according to the JEE exam syllabus. Information about The 6th term of a G.P. is 32 and its 8th term is 128 ; then the common ratio of the G.P. isa)-1b)2c)4d)-4Correct answer is option 'B'. WebThe formula for the nth term of a geometric progression whose first term is a and common ratio is r is: a n =ar n-1. The sum of n terms in GP whose first term is a and the common … jm realty management