Question

The sum of first 20 terms of an AP is 400 and sum of first 40 terms is 1600.find the sum of its first 10 terms

Answer 1

Let the first term be a and common difference is d
A/q
S20 = (20/2)[2a+(20-1)d]= 400
⇒2a +19d = 40_____(1)
also, S40 =(40/2)[2a+(40-1)d]= 1600
⇒2a +39d = 80____(2)
subtracting (1) and (2), we get
d =2
so, a= 1
S10 = (10/2)[2×1 + (10-1)2] = 5(2+18) = 5×20 = 100

Answer 2

Answer:

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.

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