Question

The 1st,3rd and 7th terms of an AP with common difference 2 are the first 3 terms of a GP.Find; a) the common ratio of the GP ganeshie8

Answer 1

@ganeshie8

Answer 2

say, the first term of AP is \(a\)

Answer 3

you're given common difference, d = 2

can you find the expressions for 3rd and 7th terms ?

Answer 4

yes

Answer 5

after that

Answer 6

first find them

Answer 7

3a+16

Answer 8

whats that ?

Answer 9

first term = \(a\)
common difference = \(2\)

3rd term = ?
7th term = ?

Answer 10

3rd term = a + 2(2)
7th term = a+6(2)

Answer 11

perfect !
so these 3 terms are in GP :
\[\large a,~ a+4, ~a+12\]

Answer 12

yes after that

Answer 13

find \(a\) using geometric mean :
\(\large (a+4)^2 = a(a+12)\)

Answer 14

can you solve \(a\) in above equation ?

Answer 15

wait

Answer 16

why you did (a+4)2=a(a+12) this can ypu please explain

Answer 17

good question :)
in GP, ratio of adjacent terms is equal, right ?

Answer 18

\[\large a,~ a+4, ~a+12\]


\(\large \dfrac{a+4}{a} = \dfrac{a+12}{a+4}\)

cross multiply and solve \(a\)

Answer 19

i know in gp ratio of adjacent terms is equalbut why (a+4)2=a(a+12)
but why (a+4)2=a(a+12)

Answer 20

wait a minute i am getting it

Answer 21

got it

Answer 22

but a+4a=a+12a+4 is AP why you used gp

Answer 23

read the question again,
the 1st, 3rd, 7th terms of given AP form first 3 terms of new GP right ?

Answer 24

yes right , did you solved this before

Answer 25

not yet, im waiting for you to finish solving it :)
let me know once you have the value of a, and the common ratio

Answer 26

i solved it , a=4 r= 2

Answer 27

looks good ^^

Answer 28

good job !!

Answer 29

Thanks genius man

Answer 30

may god make you extreme genuis