Question

The sum of first ten terms of an A.P. is four times the sum of its first five terms, then ratio of the first term and common
difference is​

Answer 1

Answer:

Fᴏʀᴍᴜʟᴀ ғᴏʀ sᴜᴍ ᴏғ AP 

$sn = \frac{n}{2} (2a + (n - 1)d)$

Sᴜᴍ ᴏғ 1sᴛ 10 ᴛᴇʀᴍs Wɪʟʟ ʙᴇ

S10. = 

$\frac{10}{2} (2a + (10 - 1)d)$

$5(2a + 9d)$

Sᴜᴍ ᴏғ 1sᴛ 5 ᴛᴇʀᴍs ᴏғ AP

S 5 = 

$\frac{5}{2} (2a + (5 - 1)d)$

$\frac{5}{2}(2a + 4d)$

In the question it is given that the sum of 1st 10 terms of AP is equal to 4 times of sum of 1st 5 terms of that AP.

S10. = 4( S5)

$5(2a + 9d) = 4 \times \frac{5}{2}(2a + 4d)$

$5(2 a+ 9d) = 10(2a + 4d)$

2a + 9d. = 2 ( 2a + 4d)

2a +9d = 4a + 8d

9d -8d. = 4a -2a

d = 2a

$\frac{a}{d} = \frac{1}{2}$

So

ᴀ:ᴅ = 1:2 

ᴡʜᴇʀᴇ ᴀ ɪs ᴛʜᴇ 1sᴛ ᴛᴇʀᴍ ᴏғ AP ᴀɴᴅ ᴅ ɪs ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ.

Answer 2

Answer:

Step-by-step explanation:

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