Question

In an A.P., the sum of the 4th and 8th term is 24 and sum of the 6th and 1Oth term is 44. Find the first 3

terms of an A.P.​

Answer 1

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Answer 2

$\underline \bold{Given \: the \: sum \: of \: 4th \: and \: 8 \: the \: term \: is \: 24}$

So ,

$\mathsf{t_8 = a + (8 -1)d}$

$\mathsf{t_8 = a + 7d}$

$\mathsf{t_4= a + (4 -1)d}$

$\mathsf{t_4= a + 3d}$

$\underline \bold{Their \: sum \: is \: 24}$

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

$\mathsf{2a + 10d = 24-----(1)}$

$\underline \bold{Given \: the \: sum \: of \: 6th \: and \: 10\: term \: is \: 44}$

$\mathsf{t_6= a + (6 -1)d}$

$\mathsf{t_6 = a + 5d}$

$\mathsf{t_10= a + (10 -1)d}$

$\mathsf{t_10= a + 9d}$

$\underline \bold{Their \: sum \: is \: 44}$

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

$\mathsf{2a + 14d =44---(2)}$

On substracting eq (2) by (1)

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

$\mathsf{-2a + 10d = 24}$

______________

$\mathsf{4d = 20}$

______________

$\mathsf{\frac{20}{4} = d}$

$\mathsf{d=5}$

So on placing value of d in any of the equation let be in eq(1).

$\mathsf{2a + 5 ×10=24}$

$\mathsf{2a + 50 =24}$

$\mathsf{2a = 24 - 50}$

$\mathsf{2a = 26}$

$\mathsf{\frac{26}{2} = a}$

$\mathsf{a=-13}$

$\mathsf{So \: a=-13 and \: d= 5}$

$\boxed {\mathsf{First \: term = -13}}$

$\boxed{\mathsf{second \: term = -13 +5 =-8}}$

$\boxed{ \mathsf{Third \: term = -8 + 5= -3}}$