Question

find five terms in AP whose sum is 25 and sum of whose squares is 135

Answer 1

Answer:

terms of AP are 3, 4, 5, 6 and 7

Step-by-step explanation:

Let a be the first term of the A.P and d be the common difference.

then,  the 5 terms be (a-2d ), (a-d) , a , (a+d) and (a+2d)

then according to question,

Sum of terms is 25.

⇒ (a-2d ) + (a-d) + a + (a+d) + (a+2d) = 25

⇒ 5a = 25

⇒ a= 5

Also given , sum of squares of terms of AP is 135

that is $(a-2d )^2+(a-d)^2+a^2+(a+d)^2+(a+2d)^2=135$

Put a = 5 and evaluate we get,

$25+4d^2-20d+25+d^2-10d+5^2+25+d^2+10d+25+4d^2+20d = 135$

Solve for d, we get,

$4d^2+d^2+d^2+4d^2= 135-125$

$10d^2=10$

$d^2=1$

$d=\pm 1$

Thus, terms are

Case 1) when a = 5 and d= 1

(a-2d ) = 5-2 = 3

(a-d) = 5-1 = 4

 a = 5

(a+d) = 5+1 = 6

and (a+2d) = 5 +2 = 7

Case 2) when a = 5 and d= -1

(a-2d ) = 5+2 = 7

(a-d) = 5+1 = 6

 a = 5

(a+d) = 5-1 = 4

and (a+2d) = 5 - 2 = 3

Thus, terms of AP are 3, 4, 5, 6 and 7

Answer 2

Step-by-step explanation:

terms of AP are 3, 4, 5, 6 and 7

Step-by-step explanation:

Let a be the first term of the A.P and d be the common difference.

then, the 5 terms be (a-2d ), (a-d) , a , (a+d) and (a+2d)

then according to question,

Sum of terms is 25.

⇒ (a-2d ) + (a-d) + a + (a+d) + (a+2d) = 25

⇒ 5a = 25

⇒ a= 5

Also given , sum of squares of terms of AP is 135

that is (a-2d )^2+(a-d)^2+a^2+(a+d)^2+(a+2d)^2=135(a−2d)

2

+(a−d)

2

+a

2

+(a+d)

2

+(a+2d)

2

=135

Put a = 5 and evaluate we get,

25+4d^2-20d+25+d^2-10d+5^2+25+d^2+10d+25+4d^2+20d = 13525+4d

2

−20d+25+d

2

−10d+5

2

+25+d

2

+10d+25+4d

2

+20d=135

Solve for d, we get,

4d^2+d^2+d^2+4d^2= 135-1254d

2

+d

2

+d

2

+4d

2

=135−125

10d^2=1010d

2

=10

d^2=1d

2

=1

d=\pm 1d=±1

Thus, terms are

Case 1) when a = 5 and d= 1

(a-2d ) = 5-2 = 3

(a-d) = 5-1 = 4

a = 5

(a+d) = 5+1 = 6

and (a+2d) = 5 +2 = 7

Case 2) when a = 5 and d= -1

(a-2d ) = 5+2 = 7

(a-d) = 5+1 = 6

a = 5

(a+d) = 5-1 = 4

and (a+2d) = 5 - 2 = 3

Thus, terms of AP are 3, 4, 5, 6 and 7