Question

The sum of 4th and 8th terms of ap is 24 and the sum of 6th and 10th term is 34. The first term is

Answer 1

Answer:

a = - 13

d = 5 / 2

Step-by-step explanation:

Let the first term a and common difference be d.

We know :

t_n = a + ( n - 1 ) d

t_4 = a + 3 d

t_8 = a + 7 d

We have given :

t_4 + t_8 = 24

2 a + 10 d = 24

a + 5 d = 12

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

t_6 = a + 5 d

t_10 = a + 9 d

: t_6 + t_10 = 34

2 a + 14 d = 34

a + 7 d = 17

a = 17 - 7 d ... ( ii )

From ( i ) and  ( ii )

12 - 5 d = 17 - 7 d

7 d - 5 d = 17 - 12

2 d = 5

d = 5 / 2

We have :

a = 12 - 5 / 2

a = 19 / 2

a = - 13

d = 5 / 2

Answer 2

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$let \: a \: be \: the \: first \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: ap \\ then \\ t _{4} + t _{8} = 24 \implies(a + 3d) + (a + 7d) = 24 \\ \implies2a + 10d = 24 \\ \implies \: a + 5d = 12 \: \: \: \: .......(1) \\ and \: t _{6} + t _{10} \implies(a + 5d) + (a + 9d) = 34 \\ \implies2a + 14d = 34 \\ \implies \: a + 7d = 17 ........(2)\\ on \: solving \: (1)and(2) \: we \: get \: a = -1/2 \: and \: d= 5/2$