Question

the sum of first 7 term of an AP is 63 and that of next 7 terms is 161 find the AP.

solve this question I will mark you brainlist answer.​

Answer 1

Answer:

The A.P is 3,5,7........

Step-by-step explanation:

Given:

• The sum of first 7 terms of an A.P = 63
• The sum of next 7 terms = 161

To Find:

• The A.P

Solution:

Here let a₁ be the first term term and d be the common difference of the A.P

The sum of n terms of an A.P is given by,

$\boxed{\tt S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)}$

where n is the number of terms

Hence by the first case given,

S₇ = 63

$\tt \dfrac{7}{2}\times (2a_1+6d)=63$

14 a₁ + 42d = 126

Dividing whole equation by 14

a₁ + 3d = 9

a₁ = 9 - 3d -------(1)

Now by second case given,

Sum of next 7 terms = 161

Therefore,

Sum of first 7 terms + Sum of next 7 terms = 63 + 161

Sum of 14 terms = 224

Hence,

$\tt \dfrac{14}{2}\times (2a_1+13d)=224$

7 × (2a₁ + 13d) = 224

Substitute value of a₁ from equation 1,

2 × ( 9 - 3d) + 13d = 32

18 - 6d + 13d = 32

7d = 14

d = 2

Hence the common difference of the A.P is 2.

Now finding the first term from equation 1,

a₁ = 9 - 3 × 2

a₁ = 9 - 6

a₁ = 3

Hence the first term of the A.P is 3.

Second term = a₁ + d = 3 + 2 = 5

Third term = a₂ + d = 5 + 2 = 7

Therefore the A.P is 3, 5, 7,.......

Answer 2

Answer:

3 , 5 , 7 ...

Step-by-step explanation:

Using S = (n/2) [2a + (n - 1)d]   ,where letters have their usual meaning.

 For first 7 terms:

⇒ 63 = (7/2) [2a + (7 - 1)d]

⇒ (63 * 2)/7 = [2a + 6d]

⇒ 18 = 2a + 6d  

⇒ 9 = a + 3d      ⇒ 9 - 3d = a

 For next 7 terms:

⇒ 161 = (7/2) [8th term + 14th term]

⇒ (161 x 2)/7 = [a + 7d + a + 13d]

⇒ 46 = 2a + 20d

⇒ 23 = a + 10d

⇒ 23 = 9 - 3d + 10d       [from above]

⇒ 2 = d

  thus, a = 9 - 3(2) = 3

∴ AP is a , a + d, a + 2d...

            3, 3 + 2, 3 + 2(2)...

            3 , 5 , 7 ....