Question

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

Answer 1

$\large\underline{Given:-}$

• Sum of 4th term and 8th term ⇢ 24
• Sum of 6th term and 10th term ⇢ 44

$\large\underline{To find:-}$

• find the first three terms of the Ap.

$\large\underline{Solutions:-}$

$\: \: \: \: \: {a_4} \: + \: {a_8} \: \: = \: \: {24}$

$\: \: \: \: \: {a} \: + \: {3d} \: + {a} \: + \: {7d} \: \: = \: \: {24}$

$\: \: \: \: \: {2a} \: + \: {10d} \: \: = \: \: {24}$

»★ Divide both side by 2.

$\: \: \: \: \: {a} \: + \: {5d} \: \: = \: \: {12} \: \: \: \: \: .....{(1)}.$

And,

$\: \: \: \: \: {a_6} \: + \: {a_{10}} \: \: = \: \: {44}$

$\: \: \: \: \: {a} \: + \: {5d} \: + {a} \: + \: {9d} \: \: = \: \: {44}$

$\: \: \: \: \: {2a} \: + \: {14d} \: \: = \: \: {44}$

»★ Divide both side by 2.

$\: \: \: \: \: {a} \: + \: {7d} \: \: = \: \: {22} \: \: \: \: \: .....{(2)}.$

✰ Subtracting Eq. (1) and Eq. (2), We get.

${a} \: + \: {5d} \: \: = \: \: {12} \\ {a} \: + \: {7d} \: \: = \: \: {22} \\ \underline{- \: \: \: \: - \: \: \: \: = \: \: - \: \: \: \: } \\ \: \: \: \: \: \: \: \: {-2d} \: \: \: \: = \: \: \: {-10}$

$\: \: \: \: \: \leadsto {d} \: \: = \: \: \frac{-10}{-2}$

$\: \: \: \: \: \leadsto {d} \: \: = \: \: {5}$

✰ Then, Putting the value of d in Eq. (1)

$\: \: \: \: \: \leadsto {a} \: + \: {5d} \: \: = \: \: {12}$

$\: \: \: \: \: \leadsto {a} \: + \: {5} \: \times \: {5} \: \: = \: \: {12}$

$\: \: \: \: \: \leadsto {a} \: + \: {25} \: \: = \: \: {12}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {12} \: - \: {25}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {-13}$

»★ Now, first three terms of Ap are a, a + d, and a + 2d.

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {-13}$

$\: \: \: \: \: \leadsto {a} \: + \: {d}\: \: = \: \: {-13} \: + \: {5} \: \: = \: \: {-8}$

$\: \: \: \: \: \leadsto {a} \: + \: {2d}\: \: = \: \: {-13} \: + \: {2} \: \times \: {5} \: \: = \: \: {-13} \: + \: {10} \: \: = \: \: {-3}$

»★Hence,

first three terms of Ap is -13, -8 And -3.

__________________________

Answer 2

Solution:

We know that, the nth term of the AP is;

an = a+(n−1)d

a4 = a+(4−1)d

a4 = a+3d

In the same way, we can write,

a8 = a+7d

a6 = a+5d

a10 = a+9d

Given that,

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.