Question

the sum of the 4th. & 8th term of an AP is 24 and the sum of 6th &10th term is 44 find the first 3 terms of AP​

Answer 1

STEP-BY-STEP EXPLANATION:

.

$As \: we \: know \: that, \begin{cases}{a}_{n} = a + (n - 1)d \end{cases}$

$\longrightarrow {a}_{4} = a + (4 - 1)d \: \\ \\ \longrightarrow {a}_{4} = a + 3d \: \: \: ...(1)$

$\longrightarrow {a}_{8} = a + (8 - 1)d \: \\ \\ \longrightarrow {a}_{8} = a + 7d \: \: \: ...(2)$

$Adding \: both \: 1 \: and \: 2,$

$\longrightarrow {a}_{4} + {a}_{8} = a + 3d + a + 7d$

$\longrightarrow 24 = 2a + 10d \: \: \:...(3) \ \: \: \: \because{a}_{4} + {a}_{8} = 24$

$Similarly,$

$\longrightarrow {a}_{6} = a + (6 - 1)d \: \\ \\ \longrightarrow {a}_{6} = a + 5d \: \: \: ...(4)$

$\longrightarrow {a}_{10} = a + (10 - 1)d \: \\ \\ \longrightarrow {a}_{10} = a + 9d \: \: \: ...(5) \: \:$

$Adding \: both \: 4 \: and \: 5,$

$\longrightarrow {a}_{6} + {a}_{10} = a + 5d + a + 9d$

$\longrightarrow 44 = 2a + 14d \: \: \:...(6) \ \: \: \: \because{a}_{6} + {a}_{10} = 44$

$Substract \: 3 \: from \: 6,$

$\longrightarrow 44 - 24= 2a + 14d - 2a - 10d$

$\longrightarrow 20= 4d$

$\longrightarrow d = 5$

$Substituting \: this \: value \: d \: in \: 3,$

$\longrightarrow 24 = 2a + 10d \: \: \:...(3)$

$\longrightarrow 24 = 2a + 10(5)$

$\longrightarrow 24 = 2a + 50$

$\longrightarrow 2a = 24 - 50$

$\longrightarrow a = - 13$

$\longrightarrow {a}_{2} = - 13 + (2 - 1)5 \: \\ \\ \longrightarrow {a}_{2} = - 13 + 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow {a}_{2} = - 8 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\longrightarrow {a}_{3} = - 13 + (3 - 1)5 \: \\ \\ \longrightarrow {a}_{3} = - 13 + 2(5) \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow {a}_{3} = - 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

REQUIRED ANSWER,

.

• The first 3 terms in AP is -13, -8 and -3.

Answer 2

QuEsTiOn,

• The sum of the 4th. & 8th term of an AP is 24 and the sum of 6th &10th term is 44 find the first 3 terms of AP.

To FiNd,

• $\color{red} \boxed{ \sf \: the \: first \: 3 \: terms \: of \: AP}$

SoLuTiOn,

$\sf \: As \: we \: know \: that, \begin{cases}{a}_{n} = a + (n - 1)d \end{cases}$

$\sf \: \longrightarrow {a}_{4} = a + (4 - 1)d \: \\ \\ \longrightarrow {a}_{4} = a + 3d \: \: \: ...(1)$

$\sf \: \longrightarrow {a}_{8} = a + (8 - 1)d \: \\ \\ \longrightarrow {a}_{8} = a + 7d \: \: \: ...(2)$

$\sf \: Adding \: both \: 1 \: and \: 2,$

$\sf \: \longrightarrow {a}_{4} + {a}_{8} = a + 3d + a + 7d$

$\sf \: \longrightarrow 24 = 2a + 10d \: \: \:...(3) \ \: \: \: \because{a}_{4} + {a}_{8} = 24$

$Similarly,$

$\sf \: \longrightarrow {a}_{6} = a + (6 - 1)d \: \\ \\ \longrightarrow {a}_{6} = a + 5d \: \: \: ...(4)$

$\sf \: \longrightarrow {a}_{10} = a + (10 - 1)d \: \\ \\ \longrightarrow {a}_{10} = a + 9d \: \: \: ...(5) \: \:$

$Adding \: both \: 4 \: and \: 5,$

$\sf \: \longrightarrow {a}_{6} + {a}_{10} = a + 5d + a + 9d$

$\sf \: \longrightarrow 44 = 2a + 14d \: \: \:...(6) \ \: \: \: \because{a}_{6} + {a}_{10} = 44$

$\sf \: Substract \: 3 \: from \: 6,$

$\sf \: \longrightarrow 44 - 24= 2a + 14d - 2a - 10d$

$\sf \: \longrightarrow 20= 4d$

$\sf \: \longrightarrow d = 5$

$\sf \: Substituting \: this \: value \: d \: in \: 3,$

$\sf \: \longrightarrow 24 = 2a + 10d \: \: \:...(3)$

$\sf \: \longrightarrow 24 = 2a + 10(5)$

$\sf \: \longrightarrow 24 = 2a + 50$

$\sf \: \longrightarrow 2a = 24 - 50$

$\sf \: \longrightarrow a = - 13$

$\sf \: \longrightarrow {a}_{2} = - 13 + (2 - 1)5 \: \\ \\ \longrightarrow {a}_{2} = - 13 + 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow {a}_{2} = - 8 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: \longrightarrow {a}_{3} = - 13 + (3 - 1)5 \: \\ \\\sf \: \longrightarrow {a}_{3} = - 13 + 2(5) \: \: \: \: \: \: \: \: \: \\ \\ \sf \: \longrightarrow {a}_{3} = - 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

FiNaL AnSwEr,

$\color{blue} \boxed{ \sf \: The \: first \: 3 \: terms \: in \: AP \: is \: -13, \: -8 \: and \: -3.}$

Hope you get your AnSwEr.