Question

If the sum of first n terms of two arithmetic progression are in the ratio (7n -5 ) : (5n + 17), show that the 6th terms of the two series are equal

Answer 1

??

Answer 2

hint: find out the value of n where the ratio of the sums of the two series are equal

Answer 3

@wwe123 do you understand?

Answer 4

ah yes !! i must admit i missed that..

Answer 5

:) it is /subtle/

Answer 6

this i known

Answer 7

so what value of n do you get for this?

Answer 8

i am not able to solve the equation

Answer 9

find the value of n that satisfies:

(7n -5 ) = (5n + 17)

Answer 10

how
(7n -5 ) = (5n + 17)

Answer 11

that would make the ratio of the sums of both series equal

Answer 12

once you have found n, then use the formula for the sum of the first n terms of a series to get a ratio of the two series at that value of n. you will then see the desired answer to the question "pop" out

Answer 13

how

Answer 14

try it yourself and you will see

Answer 15

if you list your steps here then I can guide you further

Answer 16

one min

Answer 17

\[\frac{ 2a _{1}+(n-1)d _{1} }{ 2a _{2}+(n-1)d _{2} }=\frac{ 7n-5 }{ 5n-17 }\]

Answer 18

this is right ^^^^

Answer 19

yes - but first find the value of n that makes this ratio equal to 1

Answer 20

how to find the value of n ?

Answer 21

look at the right-hand-side of your equation - set it to 1 and solve for n

Answer 22

i.e. solve this:\[\frac{ 7n-5 }{ 5n-17 }=1\]

Answer 23

but how RHS =1

Answer 24

I am postulating that if we find a value for n that makes this ratio equal to 1, then we will be able to get somewhere towards the answer.

Answer 25

ok

Answer 26

i amnot able to solve, plz show

Answer 27

surely you can solve this:\[\frac{ 7n-5 }{ 5n-17 }=1\]

Answer 28

multiply both sides by 5n-17

Answer 29

and solve for n
this is basic algebra

Answer 30

n= 11

Answer 31

good - now use this value of n in your original equation:\[\frac{ 2a _{1}+(n-1)d _{1} }{ 2a _{2}+(n-1)d _{2} }=\frac{ 7n-5 }{ 5n-17 }\]

Answer 32

to simplify it - what do you end up with?

Answer 33

there are 4 unknown value how cani solve
\[a _{1},a _{2},d _{1}, d _{2}\]

Answer 34

do not solve - just simplify the expression by substituting n=11 into it

Answer 35

\[\frac{ 2a _{1}+10d _{1} }{ 2a _{2}+10d _{2} }=\frac{ 72 }{ 38 }\]

Answer 36

@asnaseer

Answer 37

the RHS is incorrect

Answer 38

and you can divide the LHS by 2 to simplify the fraction further

Answer 39

I see the mistake - it should be 5n+17

Answer 40

not 5n-17

Answer 41

\[\frac{ 2a _{1}+(n-1)d _{1} }{ 2a _{2}+(n-1)d _{2} }=\frac{ 7n-5 }{ 5n+17 }\]

Answer 42

so ,rhs=0

Answer 43

no???

Answer 44

if n=11 what is 7n-5=?

Answer 45

sorry its 1

Answer 46

correct, so you have ended up with:\[\frac{ 2a _{1}+10d _{1} }{ 2a _{2}+10d _{2} }=1\]now simplify the fraction on the LHS (you can divide numerator and denominator by 2)

Answer 47

and then notice what each term (i.e. numerator and denominator) represent

Answer 48

i done it ...................... :)

Answer 49

good :)

Answer 50

so you see how the answer just "popped out" :)

Answer 51

yeah