If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3 (S20 − S10).
Answers
Let a be the first term and d be the common difference of the given AP.
Then sn = n/2(2a + (n-1)d).
LHS :
S30 = (30/2)(2a + (30 - 1)d)
= 15(2a + 29d)
= 30a + 435d. ---------------- (1).
RHS:
(S20 - S10) = (20/2)(2a + (20 - 1) * d) - (10/2)(2a + (10 - 1) * d)
= 10(2a + 19d) - 5(2a + 9d)
= 20a + 190d - 10a - 45d
= 10a + 145d
3(S20 - S10) = 3(10a + 145d)
= 30a + 435d ------------- (2)
Therefore From (1) and (2), It is proved that S30 = 3(S20 - S10).
LHS = RHS.
Hope this helps!
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)]
Hence Proved.
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