Question

Find 4 consecutive terms in AP whose sum is 18 and the sum of third and fourth terms is 19 please please please solve fast

Answer 1

let the terms of AP be as follows :-

$a-3d,a-d,a+d,a+3d$

Now ,

Given that :-

Sum of the AP is equal to 18

Therefore,

$a - 3d + a - d + a + d + a + 3d = 18 \\ \\ \\ 4a = 18 \\ \\ \\ a = \frac{18}{4} = 4.5$

Now,

Second part of the question says that :-

$a + d + a + 3d = 19 \\ \\ \\2a + 4d = 19 \\ \\ \\ a + 2d = \frac{19}{2} = 9.5 \\ \\ \\ 2d = 9.5 - 4.5 = 5 \\ \\ \\ d = \frac{5}{2} = 2.5$

$\rule{200}{2}$

We have :-

$a = 4.5 \\ \\ d = 2.5$

The AP is as follows :-

$a - 3d = 4.5 - 7.5 = ( - 3) \\ \\ \\ a - d = 4.5 - 2.5 = 2 \\ \\ \\ a + d = 4.5 + 2.5 = 7 \\ \\ \\ a + 3d = 4.5 + 7.5 = 12$

$\boxed{\bold{\orange{(-3),2,7,12}}}$

Answer 2

hi here is your answer dear

Check out the attachment

Ask me if any query came