Question

The sum of 5 amd 8 terms of an A.P is 37 and its 11 term is 32 find the A.P

Answer 1

Answer:-

Given:

Sum of 5th & 8th terms of an AP = 37

11th term = 32

We know that,

nth term of an AP – aₙ = a + (n - 1)d

Hence,

⟹ a₅ + a₈ = 37

⟹ a + (5 - 1)d + a + (8 - 1)d = 37

⟹ 2a + 4d + 7d = 37

⟹ 2a + 11d = 37 -- equation (1)

Similarly,

⟹ a₁₁ = 32

⟹ a + (11 - 1)d = 32

⟹ a + 10d = 32

⟹ a = 32 - 10d

Substitute the value of a in equation (1).

⟹ 2(32 - 10d) + 11d = 37

⟹ 64 - 20d + 11d = 37

⟹ 64 - 37 = 20d - 11d

⟹ 27 = 9d

⟹ 27/9 = d

⟹ 3 = d

Substitute the value of d in equation (1).

⟹ 2a + 11(3) = 37

⟹ 2a + 33 = 37

⟹ 2a = 37 - 33

⟹ 2a = 4

⟹ a = 4/2

⟹ a = 2

Now,

General form of an ap = a , a + d , a + 2d...

⟶ Required AP = 2 , 2 + 3 , 2 + 2(3)...

⟶ Required AP = 2 , 5 , 2 + 6...

⟶ Required AP = 2 , 5 , 8...

Answer 2

Given:-

• The sum of 5th and 8th terms of an A.P is 37 and 11th term is 32.

Answer:-

Using,

$\sf {a_{n}}$ = a + (n -1)d.

So as per the question,

$\sf {a_{5}}$ + $\sf {a_{8}}$ = 37.

By substituting the $\sf {a_{n}}$ formula we get,

➭[a + (5 - 1)d] +[a + (8 - 1)d] = 37

➭[a + 5d - 1d] + [a + 8d - 1d] = 37

➭a + 4d + a + 7d = 37

➭2a + 11d = 37_______ (1)

And as it is given as 11th term is 32,

So,

$\sf {a_{11}}$ = 32.

By substituting the $\sf {a_{n}}$ formula we get,

➭a + (11 - 1)d = 32

➭a + 11d - 1d = 32

➭a + 10d = 32

➭a = 32 - 10d ________ (i)

____________________________

Substituting (i) in (1):

⟹2(32 - 10d) + 11d = 37

⟹64 - 20d + 11d = 37

⟹64 - 9d = 37

⟹-9d = 37 - 64

⟹-9d = -27

⟹d = 27/9

⟹d = 3 ________ (ii)

Substituting (ii) in (1):

⟹2a + 11(3) = 37

⟹2a + 33 = 37

⟹2a = 37 - 33

⟹2a = 4

⟹a = 4/2

⟹a = 2

____________________________

So, now by using the form of an AP,

a, a + d, a + 2d we get,

⇝2, 2 + 3, 2 + 2(3)

⇝2, 5, 8.

$\bf\red{∴Required \ AP \ is \ 2, 5, 8}$