Question

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th trms is 44. Find the first three terms of the A.P.​

Answer 1

Answer:

-13, -8, -3

Step-by-step explanation:

Let the first term of an AP is a and common difference is d

according to question

T4 + T8 = 24

a + (4-1)d+ a+ (8-1)d = 24

a + 3d + a +7d = 24

(2a + 10d = 24)÷2

a + 5d = 12 ...... (i)

and

T6 + T10 = 44

a+(6-1)d + a+(10-1)d = 44

a + 5d + a + 9d = 44

(2a + 14d = 44) ÷ 2

a + 7d = 22 ....... (ii)

subtract equation (i) and (ii)

a + 5d - a - 7d = 12 - 22

-2d = -10

d = 5

from equation (i)

a + 5×5 = 12

a = 12- 25

a = -13

first three terms are -13, -8, -3

Answer 2

$\huge\underline\mathrm{Question}:-$

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

$\huge\underline\mathrm{Solution}:-$

We know, the nth term of the AP is;

an = a + (n − 1) d

an = a + (4 − 1) d

an = a + 3d

Thus, we can write,

a8 = a + 7d

a6 = a + 5d

a10 = a + 9d

Given in the question;

a4 + a8 = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12 _____________ (i)

a6 + a10 = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22 ____________ (ii)

On subtracting equation (i) from (ii), we get,

2d = 22 − 12

2d = 10

d = 5

From equation (i), we get,

a + 5d = 12

a + 5 (5) = 12

a + 25 = 12

a = −13

a2 = a + d = − 13 + 5 = −8

a3 = a2 + d = − 8 + 5 = −3

Therefore, the first three terms of this A.P. are −13, −8, and −3.