Question

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Find the sum of the two middle most terms of the given AP:
-4/3, -1, -2/3,......,13/3​

Answer 1

First term, a = - 4/3

Common difference, d = - 1 + 4/3 = ( - 3 +4 ) / 3 = 1 /3

An = a+( n - 1 ) d

13/3 = - 4 /3 + ( n - 1 ) ( 1/3 )

13/3 + 4/3 = ( n - 1 ) 1/3

17 / 3 = ( n - 1 ) 1 /3

17 = ( n - 1 )

n = 17 +1 = 18

Middle most terms = n /2 and ( n /2) + 1

=> 18/2 and ( 18 /2) +1

$\sf{\huge {=>\: 9 \:and \:10}}$

Answer 2

✧ANSWER:-✧

let the GP be

a , ar, ar^2 , ar^3, ...

the arithmetic (linear) sequence is

a , a+d , a+2d , a + 3d , a+4d ..

but 2nd term of GP= 3rd term of As

ar = a + 2d *

also for the AP

a+3d = 10 **

and 5a + 10d = 60

a + 2d = 12 ***

subtract (**) - (***)

d = -2 <------------

then in **

a - 6 = 10

a = 16 <------------

now back in *

16r = 16 -4

r = 12/16 = 3/4 <-----------

the GP is

16 , 12 , 9 , 27/4 , ...

a)

the AP is

16, 14 , 12 , 10 , 8 ,6 ...

sum(n) = (n/2)(2a + (n-1)d)

= (n/2)(32 - 2(n-1))

= (n/2)(34 - 2n)

= n(17 - n)

b) for the GP

sum(n) = a(1-r^n)/(1-r)

= 16( 1 - (3/4)^n)/(1-3/4)

= 64(1 - (3/4)^n)

sum(all terms) = a/(1-r)

= 16/(1-3/4)

= 64