Question

The sum of four numbers of an arithmetic progression is 52. The sum of their squares is 696. Find the largest term.

Answer 1

$<b><marquee>Heya</marquee>$

Let the four terms be (a - 3d), (a + 3d), (a + d) and (a - d)

A/q

Sum of these four terms = 52

a - 3d + a + 3d + a + d + a - d = 52

4a = 52

a = 13

Also,

=> (a - 3d)^2 + (a + 3d)^2 + (a - d)^2 + (a + d)^2 = 696

Using identity

(a + b)^2 = a^2 + b^2 + 2ab

and

(a - b)^2 = a^2 + b^2 - 2ab

Now,

=> (a)^2 + (3d)^2 - 2(a)(3d) + (a)^2 + (3d)^2 + 2(a)(3d) + (a)^2 + (d)^2 + 2(a)(d) + (a)^2 + (d)^2 - 2(a)(d) = 696

=> a^2 + 9d^2 - 6ad + a^2 + 9d^2 + 6ad + a^2 + d^2 + 2ad + a^2 + d^2 - 2ad = 696

=> 4a^2 + 20d^2 = 696

=> 4(a^2 + 5d^2) = 696

=> a^2 + 5d^2 = 174

=> (13)^2 + 5d^2 = 174

=> 169 + 5d^2 = 174

=> 5d^2 = 174 - 169

=> 5d^2 = 5

=> d^2 = 1

=> d = √1

=> d = -1 or +1

Case I :-

When a = 13 and d = -1

a - 3d = 13 - 3(-1) = 13 + 3 = 16

a + 3d = 13 + 3(-1) = 13 - 3 = 10

a + d = 13 + (-1) = 13 - 1 = 12

a - d = 13 - (-1) = 13 + 1 = 14

The terms are 16, 10, 12 and 14

Largest term = 16

Case II :-

When a = 13 and d = +1

a - 3d = 13 - 3(1) = 13 - 3 = 10

a + 3d = 13 + 3(1) = 13 + 3 = 16

a + d = 13 + (1) = 13 + 1 = 14

a - d = 13 - (1) = 13 - 1 = 12

Largest term = 16

Hence, the Largest term is 16

$<b><marquee>Hope this helps you.</marquee>$

Answer 2

Answer:

numbers are 10,12,14,16 and (13+3)= 16

Or, -10,-8,-6,-4.

Step by step explanation:

Let a-3d, a-d, a+d,a+3d are the four numbers.

Then sum = a-3d+a-d+a+d+a+3d = 52

4a = 52

$a = \frac{52}{4} = 13$

Sum of squares,

${(a - 3d)}^{2} + {(a - d)}^{2} + {(a + d)}^{2} + {(a + 3d)}^{2} = 696$

$4 {a}^{2} + 20 {d}^{2} = 696$

Putting value of a,

$4 \times {13}^{2} + 20 {d}^{2} = 696$

$676 + 20 {d}^{2} = 696$

$20 {d}^{2} = 696 - 676 = 20$

$20 {d}^{2} = 20$

${d}^{2} = \frac{20}{20} = 1$

$d = + 1 \: or \: - 1$

So numbers are (13-3)=10,(13-1)=12,(13+1)=14 and (13+3)= 16

numbers are (13-3)=10,(13-1)=12,(13+1)=14 and (13+3)= 16Or, -10,-8,-6,-4.