Question

Find three consecutive terms in an a.p., whose sum is 12 and the sum of whose squares is 56??

Answer 1

Answer:

The three terms in A.P. will be

2 , 4 and 6 .

Step-by-step explanation:

Let, three consecutive terms in AP be

( a - d ) , a , ( a + d )

then,

→ sum of these terms = 12

→ a - d + a + a + d = 12

→ 3 a = 12

→ a = 12 / 3

→ a = 4

and,

→ sum of there squares = 56

→ (a-d)² + a² + (a+d)² = 56

→ a² +d² -2ad +a² +a² +d² +2ad = 56

→ 3 a² + 2 d² = 56

→ 3 ( 4 )² + 2 d² = 56

→ 3 × 16 + 2 d² = 56

→ 48 + 2 d² = 56

→ 2 d² = 56 - 48

→ 2 d² = 8

→ d² = 8 / 2

→ d² = 4

→ d = √4

→ d = ± 2

So, taking d = +2 the terms will be

a - d = 4 - 2 = 2

a = 4

a + d = 4 + 2 = 6

and

taking d = -2 the terms will be

a - d = 4 - (-2) = 6

a = 4

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

Therefore,

The three terms in AP would be

2 , 4 and 6 .

Answer 2

Question:

Find three consecutive terms in an AP, whose sum is 12 and the sum of whose squares is 56??

Answer:

The three consecutive terms in A.P. are

2 , 4 and 6 .

Given:

➛There are three consecutive terms in an AP.

➛Sum of the three consecutive terms is 12.

➛Sum of squares of the terms is 56.

To Find:

The three consecutive terms.

Solution:

We are given,

➛There are three consecutive terms in an AP.

➛Sum of the three consecutive terms is 12.

➛Sum of squares of the terms is 56.

Let, three consecutive terms in AP be

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

According to given condition,

➞a - d + a + a + d = 12

➞3 a = 12

➞a = 12 / 3

⛬ a = 4

Now According to second condition,

➪ ( a - d )² + a² + ( a + d )² = 56

➪ a² + d² - 2ad + a² + a² + d² + 2ad = 56

➪ 3a² + 2d² = 56

➪3 ( 4 )² + 2 d² = 56

➪ (3 × 16 ) + 2 d² = 56

➪48 + 2 d² = 56

➪ 2 d² = 56 - 48

➪2 d² = 8

➪d² = 8 / 2

➪d² = 4

➪ d = √4

⛬ d = ± 2

When, d = +2 the terms will be

➾a - d = 4 - 2 = 2

➾a = 4

➾a + d = 4 + 2 = 6

and

When d = -2 the terms will be

➾a - d = 4 - (-2) = 6

➾a = 4

➾a + d = 4 + (-2) = 2

⛬ three terms in AP are 2 , 4 and 6 .