Question

if 4 ,x1 ,x2 ,x3 ,x4 ,x5 ,46... are in AP then find x5​

Answer 1

$a1 = 4 \\ an = 46 \\ n = 7$

$an = a + (n - 1)d \\ 46 = 4 + (7 - 1)d \\ 46 = 4 + 6d \\ 46 - 4 = 6d \\ 42 = 6d \\ d = \frac{42}{6} \\ \\ d = 7$

$x5 = a6 = a + 5d \\ a6 = 4 + 5(7) \\ a6 = 4 + 35 \\ a6 = 39 \\ a6 = x5 = 39$

Answer 2

Answer:

X5=39

hope it helps you....