Question

we have an arithmetic progression with common difference 3. given that a7+a11=-30. the value of its 13th term is

Answer 1

Step-by-step explanation:

mark me as brainliest ✌✌

Answer 2

Answér :

a(13) = 27

Note :

★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.

★ If a1 , a2 , a3 , . . . , an are in AP , then

a2 - a1 = a3 - a2 = a4 - a3 = . . .

★ The common difference of an AP is given by ; d = a(n) - a(n-1) .

★ The nth term of an AP is given by ;

a(n) = a + (n - 1)d .

Solution :

• Given : d = 3 , a(7) + a(11) = 30
• To find : a(13) = ?

We know that ,

a(n) = a + (n - 1)d ,

where a = first term and

d = common difference

Thus ,

→ a(7) = a + (7 - 1)d

→ a(7) = a + 6d

and

→ a(11) = a + (11 - 1)d

→ a(11) = a + 10d

Also ,

We have ,

→ a(7) + a(11) = 30

→ a + 6d + a + 10d = 30

→ 2a + 16d = 30

→ 2(a + 8d) = 30

→ a + 8d = 30/2

→ a + 8d = 15 ----------(1)

Now ,

The 13th term of the AP will be given as ;

→ a(n) = a + (n - 1)d

→ a(13) = a + (13 - 1)d

→ a(13) = a + 12d

→ a(13) = (a + 8d) + 4d

→ a(13) = 15 + 4×3 [ using eq-(1) ]

→ a(13) = 15 + 12

→ a(13) = 27

Hence , a(13) = 27 .