Question

five terms are in AP the difference between the square of the 5th term and the first term is 192 and the sum of second and fourth term is 16 find the first 5 terms​

Answer 1

Answer

2, 5, 8, 11, 14

$\rule{200}1$

Explanation

Let the five terms in A.P. be

(a - 4d), (a - 2d), a, (a + 2d), (a + 4d)

The first term is (a - 4d) and common difference = 2d

Given,

$\sf {T_5}^2 - {T_1}^2 = 192$

And,

$\sf T_2 + T_4 = 16$

So, we get

(a + 4d)² - (a - 4d)² = 192

And

(a - 2d) + (a + 2d) = 16

→ a - 2d + a + 2d = 16

→ 2a = 16

→ a = 8

Now,

(a + 4d)² - (a - 4d)² = 192

→ a² + 16d² + 8ad - a² - 16d² + 8ad = 192

→ 16ad = 192

→ ad = 192/16

→ ad = 12

Now, we know that a = 8

→ 8d = 12

→ d = 3/2

So, we get the terms as

(a - 4d) = 8 - 4(3/2)

→ 8 - 2(3) = 8 - 6 = 2

(a - 2d) = 8 - 2(3/2)

→ 8 - 3 = 5

a = 8

(a + 2d) = 8 + 2(3/2)

→ 8 + 3 = 11

(a + 4d) = 8 + 4(3/2)

→ 8 + 2(3) = 14

Answer 2

Answer:

Step-by-step explanation:

Given :-

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

To find :-

The five terms of A.P.

Formula to be used :-

$\bf a_n=a+(n-1)d$

Solution :-

Let the 1st term be a and common difference d.

According to the Given Condition,

,$\sf\Rightarrow(a_5)^2-(a)^2= 192\\\\\Rightarrow (a+4d)^2-a^2=192\\\\\Rightarrow a^2+(4d)^2+2 \times a\times 4d -a^2=192\\\\\Rightarrow a^2+16d^2+8ad-a^2=192 \\\\\Rightarrow16d^2+8ad=192\bf ------(i)$

Sum of second and the fourth term = 16

$\sf\Rightarrow a_2+a_4=16\\\\\Rightarrow a+d+a+3d=16\\\\\Rightarrow 2a+4d=16\\\\\Rightarrow a+2d=8 \\\\\Rightarrow a= 8-2d\bf------(ii)$

By Putting the value of a in (i), we get

$\sf\Rightarrow 16d^2+8\times(8-2d)\times d= 192\\\\\Rightarrow 16d^2+64d-16d^2=192\\\\\Rightarrow 64d=192 \\\\\Rightarrow d= \dfrac{192}{64}\\\\ \bf\Rightarrow d=3$

Putting value of d in a = 8 - 2d

We get AP as 2, 5, 8, 11, 14.

Hence, the first 5 terms​ are 2, 5, 8, 11, 14.