find the sum of first 20+40.........800
Answers
Answer:
Heya☺️✌️♥️
Step-by-step explanation:
✏️Solution :-
Let a be the first term and d be the common
difference of the given A.P.
And the sum of the first 20 terms be S(20).
S(20) = 20/2[2a + 19d]
or, 400 = 20/2[2a + 19d]
or, 400 = 10[2a + 19d]
or, 2a + 19d = 40 ..... (i)
Also, S(40) = 40/2[2a + 39d]
or, 1600 = 20[2a + 39d]
or, 2a + 39d = 80 ....(ii)
From (i) and (ii), we get
2a + 39d = 40
2a + 19d = 80
___________
- - -
⇒ 20d = 40
⇒ d = 40/20
⇒ d = 2
Putting d's value in Eq (i), we get
⇒ 2a + 19d = 40
⇒ 2a + 19(2) = 40
⇒ 2a + 38 = 40
⇒ 2a = 40 - 38
⇒ 2a = 2
⇒ a = 2/2
⇒ a = 1
Then, S(10) = [2 × 1 + (10 - 1)2]
⇒ S(10) = 5[2 + 9 × 2]
⇒ S(10) = 5[2 + 18]
⇒ S(10) = 5 × 20
⇒ S(10) = 100
Hence, the sum of its first 10 terms is 100...
✒️Hope it works♥️✌️☺️...
Step-by-step explanation:
ans:-sum of terms=16400
