Question

In an arithmetic progression U5=-1/2 and S7=21 find a, d and U9​

Answer 1

The value of a, d, and U9 is "
`\bold{ (27)/(2), - (7)/(2), \ and , - (29)/(2)}`".

Step-by-step explanation:

Given:

`U_5=-(1)/(2)\\\\S_7=21`

Find:

a, d, U9=?

formula:

`T_n=a+(n-1)d\\\\S_n=(n)/(2)(2a+(n+1)d)`

Solve:

`\to U_5 = a+4d= -(1)/(2)\\\ \to 2a+8d= -1\\\ \to 2a= -1-8d....(a)\\\\ \to S_7 =(7)/(2)(2a+(7-1)d)\\\\ \to S_7 =(7)/(2)(2a+6d)\\\\ \to 21= (7)/(2) (2a+6d)\\\\ \ put \ the \ 2a \ value\\\\ \to 21= (7)/(2) ((-1-8d)+6d)\\\\ \to 21= (7)/(2) (-1-8d+6d)\\\\ \to 21= (7)/(2) (-1-2d)\\\\ \to (21 * 2)/(7)= -1 -2d\\\\ \to 6 = -1 -2d\\\\ \to 2d=-7\\\\ \to d= -(7)/(2)`

put the value of d in equation a:

`\to 2a= -1-8 * (-7)/(2)\\\\\to 2a= -1+ 28\\\\\to 2a= 27\\\\\to a =(27)/(2)\\`

`\to \bold{U_9 = a+8d}\\\\`

put a and b value in above equation:

`\to U_9 = (27)/(2)+8*(-7)/(2) \\\\\to U_9 = (27)/(2)- 28\\\\\to U_9 = (27-56)/(2)\\\\\to U_9 = (-29)/(2)\\\\`