Question

if s10=20,s20=10 then s30=?​

Answer 1

Answer:

s30

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Answer 2

Step-by-step explanation:

Answer:

Step-by-step explanation:

Solution :-

Let a be the first term and d be the common difference of the given A.P.

Sum of the first n terms of an A.P,

Sn = n/2 [2a + ( n - 1)d]

Then, S(30) = 30/2[2a + (30 - 1)d] ....... (i) L.H.S

⇒ S[(20) - S(10)] = 20/2[2a + (20 - 1)d] - 10/2[2a + (10 - 1)d]

⇒ S[(20) - S(10)] = 10[2a + (20 - 1)d] - 5[2a + (10 - 1)d]

⇒ S[(20) - S(10)] = 10a + 190d  -  45d .

⇒ S[(20) - S(10)] = 3(10a + 145d)

⇒ S[(20) - S(10)] = 3 × 5(2a + 29d )

⇒ S[(20) - S(10)] = 30/2[2a + (30 - 1)d] ........ (ii) R.H.S

From (i) and (ii), we get

S(30) = 3[S(20) - S(10)]

= 3[10-20]

= 3[-10]

= -30