Question

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

Answer 1

Answer:

The value of a, d, and U9 is "$\bold{ \frac{27}{2}, - \frac{7}{2}, \ and , - \frac{29}{2}}$".

Explanation:

Given:

$U_5=-\frac{1}{2}\\\\S_7=21$

Find:

a, d, U9=?

formula:

$T_n=a+(n-1)d\\\\S_n=\frac{n}{2}(2a+(n+1)d)$

Solve:

$\to U_5 = a+4d= -\frac{1}{2}\\\ \to 2a+8d= -1\\\ \to 2a= -1-8d....(a)\\\\ \to S_7 =\frac{7}{2}(2a+(7-1)d)\\\\ \to S_7 =\frac{7}{2}(2a+6d)\\\\ \to 21= \frac{7}{2} (2a+6d)\\\\ \ put \ the \ 2a \ value\\\\ \to 21= \frac{7}{2} ((-1-8d)+6d)\\\\ \to 21= \frac{7}{2} (-1-8d+6d)\\\\ \to 21= \frac{7}{2} (-1-2d)\\\\ \to \frac{21 \times 2}{7}= -1 -2d\\\\ \to 6 = -1 -2d\\\\ \to 2d=-7\\\\ \to d= -\frac{7}{2}$

put the value of d in equation a:

$\to 2a= -1-8 \times \frac{-7}{2}\\\\\to 2a= -1+ 28\\\\\to 2a= 27\\\\\to a =\frac{27}{2}$

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

put a and b value in above equation:

$\to U_9 = \frac{27}{2}+8\times\frac{-7}{2} \\\\\to U_9 = \frac{27}{2}- 28\\\\\to U_9 = \frac{27-56}{2}\\\\\to U_9 = \frac{-29}{2}$