Question

If the sum of the first ten terms of an AP is four times the sum of its five terms the ratio of the first term to common difference is

Answer 1

A.P. = a, (a + d), (a + 2d), (a + 3d).....

Sum of first ten terms of an AP is four times the sum of its five terms.

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

S10 = 10/2 [2a + (10 - 1)d]

S10 = 5 (2a + 9d)

S10 = 10a + 45d

Now,

S5 = 5/2 [2a + (5 - 1)d]

S5 = 5/2 (2a + 4d)

S5 = 5/2 (2a + 4d)

S5 = 5/2 × 2(a + 2d)

S5 = 5(a + 2d)

S5 = 5a + 10d

According to question,

⇒ 10a + 45d = 4(5a + 10d)

⇒ 10a + 45d = 20a + 40d

⇒ 10a - 20a = 40d - 45d

⇒ - 10a = - 5d

⇒ 10a = 5d

⇒ 2a = d

⇒ a/d = 1/2

The ratio of the first term to common difference is 1:2

Answer 2

Answer:

1 : 2

Solution:

Let the first term of AP be 'a' and Common difference be 'd'

Using Sum of n terms of AP formula

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

Sum of first 10 terms S_10 = 10/2 [ 2a + ( 10 - 1 )d ]

⇒ S_10 = 5 ( 2a + 9d )

Sum of first 5 terms = S_5 = 5/2 [ 2a + ( 5 - 1 )d ]

⇒ S_5 = 5/2 ( 2a + 4d )

⇒ S_5 = 5/2 [ 2( a + 2d ) ]

⇒ S_5 = 5( a + 2d )

Given:

⇒ S_10 = 4 × S_5

⇒ 5( 2a + 9d ) = 4 × [ 5( a + 2d ) ]

⇒ 2a + 9d = 4( a + 2d )

⇒ 2a + 9d = 4a + 8d

⇒ 9d - 8d = 4a - 2a

⇒ d = 2a

⇒ 2a = d

⇒ a / d = 1 / 2

⇒ a : d = 1 : 2

Therefore the ratio of first term to common difference is 1 : 2.