Question

Five terms are in AP. The difference between the square of the 5th term and first term is 192
and the sum of 2nd and the 4th term is 16. Find the first 5 terms.​

Answer 1

let the AP be (a-2d),(a-d),a,(a+d),(a+2d)

ATQ,

(a+2d)^2 - (a-2d)^2 = 192

8ad =192

ad=24

(a-d) + (a+d) = 16

2a=16

a=8

ad=24

8d=24

d=3

Therefore the first five terms are-

(8-6),(8-3),8,(8+3),(8+6)

= 2,5,8,11,14

Hope this helps

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{First\:five\:terms\:of\:AP\:-2,5,8,11,14}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

Given,

• The difference between the square of 5th term and first term is 192.
• The sum of second and fourth term is 16.

To find : The first five terms of AP.

$a_{n} = a + (n - 1)d$

Now, According to the question,

${(a_{5}) }^{2 } - {(a)}^{2} = 192 \\ {(a + 4d)}^{2} - {a}^{2} = 192 \\ {a}^{2} + 16 {d}^{2} + 8ad - {a}^{2} = 192...........(i)$

Now, sum of second and the fourth term is 16.

$a_{2} + a_{4} = 16 \\ a + d + a + 3d = 16 \\ 2a + 4d = 16 \\ a + 2d = 8..........(ii) \\ a = 8 - 2d$

Substituting the value of a in (i)

We get,

$16 {d}^{2} + 8 \times (8 - 2d) \times d = 192 \\ 16{d}^{2} + 64d - 16 {d}^{2} = 192 \\ 64d = 192 \\ d = \frac{192}{64} = 3$

Now, substituting the value of d = 3 in

a = 8 - 2d

$a = 8 - 2 \times 3 \\ a = 8 - 6 \\ a = 2$

$\rule{300}{2}$

$a_{1} = 2 \\ a_{2} = a + d = 2 + 3 = 5 \\ a_{3} = a + 2d = 2 + 6 = 8 \\ a_{4} = a + 3d = 2 + 9 = 11 \\ a_{5} = a + 4d = 2 + 12 = 14$

$\rule{300}{2}$