The sum of the third and seventh terms of an A.P. s 6 and their product is 8. Find the first term and the common difference of the A.P.

Answer

`a=1, d= (1)/(2) ` OR ` a= 5, d = - (1)/(2)`

Explanation

Let the first of the A.P. be a and the common difference d.
`t_(n ) = a+ ( n-1) d ` …..(Formula)
`:. t_(3) = a+ ( 3-1) d `
`:. T_(3) = a+ 2d` …(1)
Similarly , `t_(2) = a+ 6d` …(2)
Now, `t_(3)+t_(7) = 6` …(Given )
`:. a + 2d a+6d = 6` ....[From (1) and (20 ]
`:. 2a + 8d = 6 ` `:. a + 4d = 3 ` .... (Dividing both the sides by 2 )
`:. a = 3-4d ` ....(3)
`t_(3) xx t_(7) = 8` ....(Given )
`:. ( a + 2d) ( a+ 6d) = 8`
Substituting `a= 3-4d ` from (3)
`( 3-4d + 2d) ( 3-4d+6d) = 8 `
`:. ( 3-2d ) (3+ 2d)= 8` `:. 9-4d^(2) = 8` `:. 9-8 = 4d^(2)`
`:. 4d^(2) = 1 ` `:. d^(2) = (1)/(4)` `:. d = +- (1)/(2)`
Substituting `d= (1)/(2) `in equation (3),
`a= 3-4((1)/(2)) = 3-2` `:. a = 1 `.
Substituting `d= - (1)/(2)` in equation (3) .
`a= 3-4 (-(1)/(2)) = 3+2 ` `:. a = 5 `