Question

If S20 = S40 then find A.P of S60

Answer 1

Answer:

850

Step-by-step explanation:

hope this can be help

Answer 2

Answer: S60 = 0

Step-by-step explanation:

We know that,

$S_{n} = \frac{n}{2}(2a_{1} + (n - 1)d)$

where, S is the sum

$a_{1}$ = first term

n = number of terms

d = common difference

∴ $S_{20} = \frac{20}{2}(2a_{1} + (20 - 1) d )$

⇒ $S_{20} = 10(2a_{1} + 19 d )$

∴ $S_{40} = \frac{40}{2}(2a_{1} + (40 - 1) d )$

⇒ $S_{40} = 20(2a_{1} + 39 d )$

Given, $S_{20} = S_{40}$

⇒ $10(2a_{1} + 19 d) = 20(2a_{1} + 39 d)$

⇒ $2a_{1} + 19d = 4a_{1} + 78d$

⇒ 2a = - 59d

∴ $S_{60} = \frac{60}{2}(2a + 59d)$

         = 30(-59d + 59d)

         = 0  (Answer)