Question

In an ap Fourth term Is -13/2, seventeenth zero And last One is 21. find the first term

Answer 1

Answer:

the first term is -8

Step-by-step explanation:

$t_{4} = \frac{-13}{2}$

$a + (n - 1)d = \frac{-13}{2}\\\\a+(4-1)d= \frac{-13}{2}\\\\a = \frac{-13}{2} - 3d$

$a = \frac{-13 - 6d}{2}$                   ...............(equation 1)

$t_{17} = 0$

$a + (n-1)d = 0\\\\a +(17-1)d=0$

$\frac{-13 - 6d}{2} +16d=0$                      (from equation 1)

$\frac{-13 - 6d + 32d}{2}=0\\\\-13 + 26d = 0\\\\26d = 13\\\\d = \frac{13}{26}\\\\d = \frac{1}{2}$

now putting the value of d in equation 1

$a = \frac{-13 - 6(\frac{1}{2})}{2}\\\\a = \frac{-13 - 3 }{2} \\\\a = \frac{-16}{2}\\\\a = -8$

I hope it helps......