Five terms are in AP. The difference between the square of the 5th ter

1.	Five terms are in AP. The difference between the square of the 5th term and first term is 192and the sum of 2nd and the 4th term is 16. Find the first 5 terms.
Answer» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\tt{\therefore{First\:<klux>FIVE</klux>\:terms\:of\:AP\:-2,5,8,11,14}}}$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ 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, $16 {d}^{2} + 8 \times (8 - 2d) \times d = 192 \\ 16{d}^{2} + 64d - 16 {d}^{2} = 192 \\ 64d = 192 \\ d = \frac{192}{64} = <klux>3</klux>$ Now, substituting the value of d = 3 in a = 8 - 2d $a = 8 - 2 \times 3 \\ a = 8 - 6 \\ a = 2$ $\rule{300}{2}$ $a_{1} = 2 \\ a_{2} = a + d = 2 + 3 = 5 \\ a_{3} = a + 2d = 2 + 6 = 8 \\ a_{4} = a + 3d = 2 + 9 = 11 \\ a_{5} = a + 4d = 2 + 12 = 14$ $\rule{300}{2}$

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

Answer»

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

$\green{\tt{\therefore{First\:<klux>FIVE</klux>\:terms\:of\:AP\:-2,5,8,11,14}}}$

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

Given,

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,

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

Now, substituting the value of d = 3 in

a = 8 - 2d

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

$\rule{300}{2}$

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

$\rule{300}{2}$