Question

Plz help!!! In an arithmetic series t1 = 11/3 , t3 = 3. Find the value of n for which Sn = 20.

Answer 1

Answer:

8 and 15

Step-by-step explanation:

Given=  

T1 = 11/3    T2 = ?     T3 = 3           Sn= 20

Here , a(First term) = 11/3

and t2 = arithmatic mean of T1 and T3 = (11/3 + 3) / 2 = 10/3

Now, T2 = 10/3

From T1 and T2 we have "d" (common difference) = T2 - T1 = 10/3 - 11/3 = -1/3

Here , d = -1/3

As we know -

Sn= n/2 { 2a + d(n-1)} = 20

20=n/2[{ 2(11/3) + (-1/3)(n-1)}]

20*2 = n(22/3 - n/3 + 1/3)

40 = n( 23/3 - n/3 )

40 = 23n/3 -n^2/3

40 = ( 23n - n^2 )/3

120 = 23n - n^2

n^2 -23n +120 = 0

Now , By factorization

n^2 - 8n - 15n +120

n(n-8) -15(n-8)

(n-8) (n-15)

So, n=8

and n= 15

Hope it'll help  ...

Thanks..