Question

find the sum of first 20+40.........800​

Answer 1

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♥️✌️☺️...

Answer 2

Step-by-step explanation:

ans:-sum of terms=16400