Question

Let An be the nth term of an A.P. Let Sn be the sum of first n terms of the A.P with A1= 1 and A3 = 3)(A8). Find the largest value possible of Sn.​

Answer 1

Answer:

I don't know the answer sorry...

Answer 2

Answer:

largest possible value of $S_{n}$ =$\frac{100}{19}$

Step-by-step explanation:

Given:-   $a_{1}$ = A = 1        $a_{3}$ = 3$a_{8}$    

$T_{n}$ = A+ (n -1)d

$a_{3}$ = A +2d

$a_{8}$ = A+ 7d

A+ 2d = 3(A + 7d)

-2A = 19d

d = $\frac{-2}{19}$

$S_{n}$ = $\frac{n}{2}$ [2A+ (n-1)d]

    = $\frac{n}{2}$ [2 +(n-1)-2/19]

    = n+ $\frac{n}{2}$ (n-1)(n-2)/19

    = n - n(n-1)/19

$S_{n}$ = $\frac{20n -n^{2} }{19}$

For maxima;

$\frac{dS_{n}}{dn}$ = $\frac{1}{19}[20 - 2n]$ = 0

     ⇒ $20 - 2n$ = 0

     ⇒ 2n = 20

     ∴ n = 10

$\frac{d^{2} S_{n} }{dn}$ = $\frac{-2}{19}$ < 0

∴ $S_{n}$ =  $\frac{20n -n^{2} }{19}$

       = $\frac{20(10) - (10)^{2} }{19}$

       = $\frac{200 - 100}{19}$

       = $\frac{100}{19}$