Question

the sum of four numbers of an ap is 52 and sum of their squares is 696 find the largest term

Answer 1

do you mean alternate route if yes then the answer will be

x+x+2+x+4+x+6=52
4x+12=52
x = (52-12)/4
x=10
therefore the biggest term is 16

hope it helps plz Mark as brainliest

Answer 2

Answer:

The largest term is 16

Step-by-step explanation:

Given:

4a = 52;

a= 52/4 =13.

Sum of their squares =>$(a-3 d)^{2}+(a+3 d)^{2}+(a-d)^{2}+(a+d)^{2}$=696

Step 1:

Using identity

$(a+b)^{2}=a^{2}+b^{2}+2 a b$

and

$(a-b)^{2}=a^{2}+b^{2}-2 a b$

Step 2:

Now,

$\begin{array}{l}{\Rightarrow(a)^{2}+(3 d)^{2}-2(a)(3 d)+(a)^{2}+(3 d)^{2}+2(a)(3 d)+(a)^{2}+(d)^{2}+2(a)(d)+(a)^{2}+} \\ {(d)^{2}-2(a)(d)=696} \\ {\Rightarrow a^{2}+9 d^{2}-6 a d+a^{2}+9 d^{2}+6 a d+a^{2}+d^{2}+2 a d+a^{2}+d^{2}-2 a d=696}\end{array}$

Step 3:

$\begin{array}{l}{=>4 a^{2}+20 d^{2}=696} \\ {=>4\left(a^{2}+5 d^{2}\right)=696} \\ {=>a^{2}+5 d^{2}=174} \\ {=>(13)^{2}+5 d^{2}=174} \\ {\Rightarrow 169+5 d^{2}=174} \\ {=>5 d^{2}=174-169} \\ {=>5 d^{2}=5} \\ {\Rightarrow d^{2}=1}\end{array}$

=> d = √1

=> d = -1 or +1

Step 4:

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

Step 5:

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